Introduction

In this chapter, we will study three curves: the parabola, the ellipse and the hyperbola,
CHAP 5: CONICS which are all known as conic sections or conics.
In coordinate geometry, a conic is the locus of a point which moves so that its distance
from a fixed point is a constant ratio to its distance from a fixed line.

• The fixed point is the focus of the conic

Objectives: Contents:
• The fixed line is the directrix, and
• Define a parabola • The constant ratio is the eccentricity and is
• Determine the Cartesian and parametric equations of a denoted by e
 e, the locus of point P is a conic.
parabola FP

When: e  1, the conic is a parabola,

• Find out the tangents, axis of symmetry of a parabola 5.1. The parabola: PM
• 5.2. The ellipse •
e  1, , the conic is an ellipse,
Define an ellipse
• 5.3. The hyperbola
e  1, the conic is a hyperbola.
Determine the Cartesian and parametric equations of
an ellipse 5.4. Conics and eccentricity
• Find out the tangents, axis and centre of symmetry of 5.5. Conics and polar coordinates
an ellipse
• Determine the nature and eccentricity of a conic
• Convert polar coordinates in Cartesian coordinates and
vice versa
• Determine the equations of conics in polar coordinates

5.1. The parabola:

Prerequisite knowledge
(1) Distance formulas 5.1.1 Definition: focus and directrix of a parabola
The distance between two distinct points A(x1 , y1 ) and B(x2 , y2 ) is

given by dA, B  (x2  x1 )2  (y2  x2 )2

(2) The distance from the point P(x0 , y0 ) to the line l ≡ ax  by  c  0
Let the distance from the fixed point F (focus) to
|ax by c|
dP, l 
the fixed line D (directrix) be 2a, then
is given by • The coordinates of F are (a, 0)
a2 b2
• The equation of the directrix is
D ≡ x  a or D ≡ x  a  0.
Cartesian plane subject to the condition y  f(x).
(3) A curve or a graph of a function f is the set of all points on the

(4) The angle between two lines l1 ≡ y  m1 x  c1 and l2 ≡ y  m2 x  c2 is

defined by
tanθ  1m m2
m m
1 2

Prepared by Jean Paul RUBAYIZA Page 225 Prepared by Jean Paul RUBAYIZA Page 226
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
5.1.2 Cartesian and parametric equations of a parabola Let P(x ,y) be any point on the parabola.

From the definition of a conic section, PM  e

PF PF=(x  h  a)2  (y  k)2 , PM  |x  h  a|

1 ⇔ PF  PM,
PF=(x  a)2+(y  0)2 ,
PF
For a parabola, PM

PM  |x  a| (x  h  a)2  (y  k)2  |x  h  a|

For a parabola, e  1
∴  1 ⇔ PF  PM
Squaring boyh sides:
(x  h  a)2  (y  k)2  (x  h  a)2
PF
PM

(x  a)  (y  0)  |x  a|
2 2
x2  h2  a2  2hx  2ax  2ah  y2  2ky  k2  x2  h2  a2  2ax  2hx  2ah
y2  2ky  k2  4ax  4ah
(x  a)2  (y  0)2  (x  a)2
Squaring both sides:
(y  k)2  4a(x  h)
x2  2ax  a2  y2  x2  2ax  a2 ,
y2  4ax Other standard forms are:
• The graph of this equation is symmetric with respect to the x -axis.
• The point in which the curve crosses the axis of symmetry is called the vertex. (y  k)2  4a(x  h)
• The chord CC' through the focus and perpendicular to the axis is called the (x  h)2  4a(y  k)
latus rectum (LR). The length of the latus rectum is 4a (LR  4a), the
coefficient of the first degree term. When expanded, the equations take the form
y  ax2  bx  c
x  ay2  by  c
Remarks

y2  4ax
1. If the focus was to the left of the directrix, the equation would have the form:

Note: The equation of the parabola y2  4ax can be represented parametrically by

x2  4ay
2. If the focus was on the y-axis, the form of the equation would be

x  at2
 , t∈
y  2at
the sign depending on the position of the focus above or below the directrix

Consider now a parabola with:

From the second equation: t  2a
y

Substitute t  2a in the first equation:

• Its vertex at point (h, k), y
• Axis parallel to the x – axis, and
x  a 
2
• Focus at a distance a units to the right of y

x  4a2
2a

Let the distance from the focus F(h  a, k)

the vertex. ay2
•
to directrix d ≡ x  h  a be 2a x
y2

y2  4ax as required.
4a

Prepared by Jean Paul RUBAYIZA Page 227 Prepared by Jean Paul RUBAYIZA Page 228
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex1: Given the parabola with equation y2  4x. Exercise 5.1

(a) focus at (3, 0) and the directrix is x  3  0.

(a) Locate the vertex and focus of the parabola. (b) Find the equation of the directrix. 1. Derive the equation of each of the following parabolas:

(b) focus at (0, 6) and the directrix is the x–axis (y  0).

Sln: (a) y2  4x
(c) Determine the length of the latus rectum. (d) Hence plot the given parabola
(c) vertex at the origin, axis along x–axis, and passes through (  3, 6).
4a  4 ⇔ a  1 (d) axis vertical, and passes through the points (4, 5), (  2, 11) and (  4, 21).
(d) Curve sketching.

(e) focus at (  1,  2). and directrix x  2y  3  0.

∴ V  (0, 0) 2. Given the parabola with equation x2  4y  0,
Vertex is at the origin and the axis

F  (a, 0)  (1, 0)
The axis of symmetry is the x-axis , so (a) Locate the vertex and focus of the parabola (b) Find the equation of the directrix
(c) Determine the length of the latus rectum (d) Plot the given parabola

d ≡ x  a
(b) Equation of directrix
5.1.3 Tangents and axis of symmetry
∴ d ≡ x  1

LR  4a
(c) Length of the latus rectum (a) Cartesian equation of the tangent line at a given point of the parabola

LR  4(1)  4 Derive the equation of the tangent to the parabola y2  4ax at the point P(x0 , y0 ).
The equation of the tangent line is given by

y  y0  m(x  x0 ) (1) y0 y  y20  2ax  2ax0

where m  f ' x0 .
(2)

y  4ax ⟹ 2yy'  4a parabola y2  4ax then

But if the point P(x0 , y0 ) is to lie on the

y20  4ax0.
2

y'  y
2a

∴ At P(x0 , y0 ); m  y0 y  4ax0  2ax  2ax0

Ex2: Derive the equation of the parabola with its axis parallel to the x – axis, and which passes 2a Introducing this relation into (2) we get :

y0 y  2ax  2ax0
through the points (2, 1), (1, 2) and (1, 3). y0

x  ay2  by  c a  2.5, b  10.5, c  10 ∴ T ≡ y0 y  2ax  x0 

Equation (1) becomes
y  y0  (x  x0 )
Sln: The desired equation is in the form Solving these simultaneous equations:
2a

∴ x  2.5y2  10.5y  10
y0

2x  5y2  21y  20
Substituting for x and y the

a  b  c  2
Remark
5y2  2x  21y  20  0.
coordinates of the points;

 4a  2b  c  1 The perpendicular line to the tangent at the point of tangency is called the normal line
9a  3b  c  1 and it is given by y  y0   (x  x0) where m  f' x0 .
1

Ex3: Write the equation of the parabola with its focus at the point (6, 2) and whose
m

Ex1: Find equations of the tangent and normal lines to the parabola y  4x2 at the point (  1, 4).
directrix is the line x  2.
Sln: y  4x2 ⟹ y'  8x
Sln: x  2 ⇔ x  2  0
∴ (x  6)2  (y  2)2  |x  2| At (  1, 4), the slope of the tangent line is
m  8(  1)  8
If P(x, y) is any point on the
N ≡ y  y0   (x  x0 )
The normal line is given by

PF  PM T ≡ y  y0  m(x  x0 )
parabola, then 1

(x  6)2  (y  2)2  (x  2)2

Squaring both sides:
y  4  8 (x  1)
m

 12x  36  y2  4y  4  x2  4x  4 y  4  8(x  1)
PF  (x  6)  (y  2) ,
1

y2  4y  8x  36  0 y  4  8x  8 8y  32  x  1
2
2 2 x

PM  2 2  |x  2|
|x 2|
∴ T ≡ 8x  y  4  0 x  8y  33  0
1 0 ∴ N ≡ x  8y  33  0

Prepared by Jean Paul RUBAYIZA Page 229 Prepared by Jean Paul RUBAYIZA Page 230
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex2: Find the equation of the tangent line to the parabola y2  8x which is
parallel to the line 2x  2y  3  0. x = 2t2
Ex5: Find the equation of the tangent to the parabola

y = 4 + 4t
Sln: 2x  2y  3  0 ⟺ y  x  at the point where t = −1.
For y0 = −4, y0 2 = 8x0 Slope of the tangent line when t = −1:
3

m1  1
x = 2( − 1)2 = 2 m = = = −1
2

8x0 = ( − 4)2
Sln: Coordinates of the point of tangency P
 ∴ P(2, 0)
y = 4 + 4( − 1) = 0
dy 4

y2  8x ⟹ 2yy'  8 8x0 = 16
Slope of the tangent at P(x0 , y0 )
x'(t) = 4t
dx 4

x0 = 2  T ≡ y − y0 = m(x − x0 )
The tangent line at P(2, 0) is given by
y'  T ≡ y − y0 = m(x − x0 ) y'(t) = 4
y − 0 = −1(x − 2)
4

x'( − 1) = −4
∴ At P(x0 , y0 ); m2 = y y + 4 = −1(x − 2) For t = −1,  y = −x + 2
y

y + 4 = −x + 2 y'( − 1) = 4
4

x+y+4−2= 0
0

line 2x + 2y − 3 = 0, m1 = m2
Since the tangent line is parallel to the
x + y + 2 = 0.
= −1 ⟺ y0 = −4
(b) Axis of symmetry of a parabola

If the standard form of the parabola is (y − k)2 = 4a(x − h), then its axis of symmetry is
4

parallel to the x–axis and its equation is y − k = 0.

y 0

Ex3: For what value of m the line y = mx + 2 is tangent to the parabola y2 = 4x?

If the standard form of the parabola is (x − h)2 = 4a(y − k), then its axis of symmetry is
y = mx + 2 parallel to the y–axis and its equation is x − h = 0.

y2 = 4x ∆= (4m − 4)2 − 16m2 = 0
(1)
Sln:
Replace y by mx + 2 in (2): 16m2 − 32m + 16 − 16m2 = 0
(2)
Ex1: Write the equation of the parabola with its vertex at the point (2, 3), its axis
(mx + 2 )2 = 4x −32m + 16 = 0
parallel to the y – axis, and which passes through the point (4, 5).
m2 x2 + 4mx + 4 = 4x m=2
(x − h)2 = 4a(y − k)
Sln: the standard form to be used is
m2 x2 + 4mx − 4x + 4 = 0
1

∴ m = and (x − 2)2 = 4a(y − 3) (x − 2)2 = 4 × (y − 3)

The required equation is
m2 x2 + (4m − 4)x + 4 = 0
1 1

T≡y= +2 x − 4x + 4 = 2y − 6
2 2

(4 − 2) = 4a(5 − 3)
x

x2 − 4x − 2y + 10 = 0
To have a point of tangency, the since the point (4, 5) is on the curve; 2

8a = 4
2 2
discriminant must be equal to zero.

Ex4: One of the points of intersection of the curves y2 = 4x and 2x2 = 12 − 5y is (1, 2). a = 1/2
Find the acute angle of intersection of the curves at that point.

Sln: The acute angle of intersection between two

tanθ = =  =  51 
4 9
curves is defined to be the angle between their m1 m2 1

5

(1) y2 = 4x ⟹ 2yy' = 4
tangents at their intersection point.
=9
4
1 m1 m2 1 1
5 5

y' = y ∴ θ = arctan9 = 83.7°

2

At (1, 2) , m1 = 2 = 1
2

(2) 2x2 = 12 − 5y ⟹ 4x = −5y'

y' = − 5
4x

At (1, 2) , m2 = − 5
4

Prepared by Jean Paul RUBAYIZA Page 231 Prepared by Jean Paul RUBAYIZA Page 232
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex2: The equation of a parabola is x2  2x  2y  3  0. 5.1.4 Characteristic elements of a parabola

(y  k)2  4a(x  h) (x  h)2  4a(y  k)

(a) Write the equation of the parabola in standard form and determine:
(i) the coordinates of the vertex Equation
(ii) the coordinates of the focus
(iii) the equation of the directrix (b) latus rectum 4a 4a
(b) Sketch the parabola on the Cartesian plane. Vertex V(h, k) V(h, k)
F(h  a, k) F(h, k  a)
x2  2x  2y  3  0
Sln: (a) Completing the squares
yk xh
Focus

x2  2x  2y  3
Focal axis or

x  2x  1  2y  3  1 D≡xha D≡yka
Axis of symmetry
2 Directrix
(x  1)2  2y  4
(x  1)2  2 y  2
(i) Vertex: h  1, k  2
∴ V  (h, k)  (1,  2)
(ii) Axis of symmetry // to the y–axis
a0
4a  2 ⇔ a  0.5
F  (h, k  a)  (1,  2  0.5)  (1,  1.5)
(iii) Equation of directrix: d ≡ y  k  a Curve
∴ d ≡ y  2  0.5
d ≡ y  2.5

Exercise 5.2
1. Write the equations of the tangent and normal to each of the following parabola at the

(a) x2  4y  0 at the point (2, 1). a0

indicated point.

(b) y2  x at the point (1, 1)

(e) 4(x  1)2  y at the point (  3, 16)
2. For what value of m the line y  mx  1 is tangent to the parabola y2  12x?

x  2  2t
3. The parametric equations of a curve C are:

y  4  t2 Exercise 5.3
(a) Find the Cartesian equation of the curve (C). Which shape is it?
from (  2, 3) equals its distance from the line x  6  0.
1. (a) Derive the equation of the locus of a point which moves so that its distance
where t  1.
(b) Determine the equation of the tangent line to the curve C at the point

through the points (3, 3), (6, 5) and (6,  3).

(b) Find the equation of a parabola with its axis parallel to the x–axis, and which passes
(c) Find its characteristic elements and represent it graphically.
2. Reduce the equation of each of the following parabolas to the standard form, and find the
coordinates of
(a) the vertex, (b) the focus, (c) the length of the latus rectum,

(1) y2  4y  6x  8  0
(d) the equation of the directrix.

(2) 4y  x2  16
(3) y2  4y  6x  13  0

Prepared by Jean Paul RUBAYIZA Page 233 Prepared by Jean Paul RUBAYIZA Page 234
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
3. For what point of the parabola y2  18x, is the ordinate equal to three times the abscissa?
4. (a) Derive the equation of the parabola with its focus at (  2,  1), and whose latus
5.2 The ellipse

rectum joins the points (  2, 2) and (  2,  4).

(b) Find the equation of the parabola with its vertex on the line 2y  3x  0, its axis
5.2.1 Definition: foci and directrices of an ellipse

parallel to the x–axis, and which passes through the two points (3, 5) and (6,  1).

(a) y2  6y  8x  31  0 at the point (  3,  1)..

5. Write the equation of the tangent and normal lines to each of the following parabolas An ellipse is defined as the path of a point which

(b) x2  5y at the point whose abscissa is 3.

moves so that the sum of its distances from two
fixed points is a constant. The two fixed points are

x  t2  t
6. A curve is defined by the parametric equations foci of the ellipse.

y  3t  4 •
F1 (c, 0) and F2 (  c, 0)
The coordinates of the two foci are:
Find the equation of the tangent to the curve at (2, 10)
•
x2  4y  16  0
7. Given a parabola (P) with equation: The equations of the directrices d1 and d2 are

d1 ≡ x  e and d2 ≡ x   e .
given by
a a
Required:
(a) Write the equation of the parabola (P)to the standard form
(b) Find the coordinates of the vertex, and the focus of the parabola (P)..

(d) On the same graph, plot the line d with equation x  y  1  0

(c) On the rectangular coordinate system, sketch the parabola (P).

(e) Use your graph to find the points of intersection of the parabola (P) with the line (d)
Let the constant sum be 2a, (a  c) and let P(x, y) be any point on the ellipse.
5.2.2 Cartesian and parametric equations of an ellipse

From the definition, F1 P  F2 P  2a (1)

F1 P  (x  c)2  (y  0)2

F2 P  (x  c)2  (y  0)2

Equation (1) becomes:

(x  c)2  y2  (x  c)2  y2  2a

(x  c)2  y2  2a  (x  c)2  y2

Squaring: (x  c)2  y2  4a2  (x  c)2  y2  4a(x  c)2  y2

x2  2cx  c2  y2  4a2  x  2cx  c2  y2  4a(x  c)2  y2

2

4cx  4a2  4a(x  c)2  y2

cx  a2  a(x  c)2  y2
Squaring again: c2 x2  2ca2 x  a4  a2 (x  c)2  y2 !
c2 x2  2ca2 x  a4  a2 "x2  2cx  c2  y2 #
c2 x2  2ca2 x  a4  a2 x  2ca2 x  a2 c2  a2 y2
2

Prepared by Jean Paul RUBAYIZA Page 235 Prepared by Jean Paul RUBAYIZA Page 236
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 a2x  c2 x2  a2 y2  a4  a2 c2
2

(a2  c2 )x2  a2 y2  a2 (a2  c2 )

Dividing each term by a2(a2  c2 ), the equation becomes
5.2.3. Characteristic elements of the ellipse with center at (h, k)

x2
 a2
y2
1 ab
 1  1
Since a  c, a  c  0. Let a  c  b , then we have the standard form of the equation of an
a2 c2 Equation (x h)2 (y k)2 (x h)2 (y k)2
2 2 2 2 2 2 2

c a2  b
a2 b b a2
Focal length
 21 ⟺ b2 x2  a2 y2  a2b2
2
ellipse,
e 
x2 y2
2 a b
Eccentricity c a2 b2
a a

LR 
If the center was at the point (h, k) the equation would be Length of latus 2b2

V1(h  a, k) V1 (h  b, k)
rectum
 1
(x h)2 (y k)2
a

V2 (h  a, k) V2 (h  b, k)
Vertices / Most (i) Left most point: (i) Left most point:
a2 b2

V3 (h, k  b) V3 (h, k  a)
points (ii) Right most point: (ii) Right most point:

(iv) Bottom most point: V4 (h, k  b) (iv) Bottom most point: V4 (h, k  a)
This equation can be represented parametrically by: (iii) Top most point: (iii) Top most point:
x  h  acosθ

y  k  bsinθ
Axes of symmetry x  h, y  k x  h, y  k
 cosθ  cos2 θ
(x h)2

⟹  F1, 2 (h  c, k) F1, 2 (h, k  c)

x h

 bsinθ
a2

 sin2 θ
a Recall that: Foci
cos2 θ  sin2 θ  1
In fact:
d1, 2 ≡ x  h  d1, 2 ≡ y  k  e
y k (y k)2
b 2 a a
b Directrices
 1
2 2
e
(x h) (y k)
a2 b2
In particular, the parametric equations of the ellipse centered at the origin are
x  acosθ

y  bsinθ
Curve

  1 is expanded, it takes the form:

(x h)2 (y k)2
When the equation a2 b2

Ax2  By2  Cx  Dy  F  0
where A and B have the same sign

Prepared by Jean Paul RUBAYIZA Page 237 Prepared by Jean Paul RUBAYIZA Page 238
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex1: For the ellipse 25x2  9y2  225, find (b) a2  36, b2  16 and so a  6, b  4

V1, 2  (h  a, k)  (6  6,  4): V1  (0,  4) , V2  (12,  4)

(a) the semi – major axis and semi – minor axis, (c) The vertices are:

V3, 4  (h, k  b)  (6,  4  4): V3  (6, 0) , V4  (6,  8)

(b) the coordinates of the foci,
(c) the eccentricity,
(d) c  a2  b2  √36  16  √20  2√5
(d) the equations of the directrices,

F1, 2  (h  c, k)  (6  2√7,  4)
(e) the length of the latus rectum and plot its curve.

Sln: (a) Given the equation 25x2  9y2  225 ∴ F1  (6  2√5,  4)  (10.5,  4)
Divide both sides by 225: Curve sketching F2  (6  2√5,  4)  (1.5,  4)
 1 (e) The length of the latus rectum: LR     5.3
x2 y2 2b2 216 16

By identification a  25, b  9, so
9 25 a 6 3
2

a  5 (semi- major axis)

2

b  3 ( semi- minor axis)

(b) Foci: F1, 2  (0,  c)
c  a2  b2  √25  9  √16  4
∴ F1, 2  (0,  c)  (0,  4)
(c) Eccentricity: e  b  b  5
c a2 b2 4

d1, 2 ≡ y   ;
(d) The equations of the directrices are
a

 
e
a 5 25
4
e 4

∴ d1, 2 ≡ y  
5
25
4
(e) The length of the latus rectum:
LR    3.6
2b2 29
a 5

Ex.3: A point moves so that the sum of its distances from the points (4, 2) and (2, 2) is 8.

Ex2: Given the ellipse 4x2  9y2  48x  72y  144  0,

Derive the equation of its locus.

F1 P  F2 P  8, where P(x, y) is any point on the locus

Ans.
(a) Find its center (b) Find its semi – axes (c) Find its vertices,
(x  4)2  (y  2)2  (x  2)2  (y  2)2  8
(d) Locate its foci (e) Determine the length of the latus rectum and plot its curve.

Sln: (a) Given the 4x  9y  48x  72y  144  0

(x  2)2  (y  2)2  8  (x  4)2  (y  2)2
2 2

4x2  12x  9y2  8y  144  0

Completing the squares:

4x2  12x  36  36  9y2  8y  16  16  144  0 (x  2)2  (y  2)2  64  (x  4)2  (y  2)2  16(x  4)2  (y  2)2
4x2  12x  36  9y2  8y  16  144  144  144  0
4(x  6)2  9(y  4)2  144 x2  4x  4  64  x2  8x  16  16(x  4)2  (y  2)2

 16  1 12x  76  16(x  4)2  (y  2)2

2 2
(x 6) (y 4)

By identification h  6, k  4 and so C(6,  4)

36

3x  19  4(x  4)2  (y  2)2

Prepared by Jean Paul RUBAYIZA Page 239 Prepared by Jean Paul RUBAYIZA Page 240
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 9x2  114x  361  16 (x  4)2  (y  2)2 ! 1. Given the ellipse 9x2 + 16y2 − 36x + 96y + 36 = 0,
Squaring both sides: Exercise 5.4

9x2  114x  361  16x2  8x  16  y2  4y  4 (a) Find the coordinates of its center, (b) Find its semi – axes,
9x2  114x  361  16x2  128x  16y2  64y  320
(c) Find its vertices, (d) Locate its foci,
7x2  16y2  14x  64y  41  0
2. Find the equation of the ellipse with its focus at (4, − 3), directrix x = −1, and
(e) Determine the length of the latus rectum (f) Plot its curve.
(ellipse)

3. A point P(x, y) moves so that the sum of its distances from the points (3, 1) and ( − 5, 1)
eccentricity 2/3.
the line x  16  0. Find the equation of its locus.
Ex. 4: A point moves so that its distance from the point (4, 0) is always one half its distance from
is 10. Derive the equation of its locus. What curve is it?
Sln: F(4, 0), l ≡ x − 16 = 0
∴ (x − 4)2 + y2 = 2 |x − 6| points ( − 2, 1) and (6, 5) is −4. Show that the locus is an ellipse and locate its center.
4. A point P(x, y) moves so that the product of the slopes of the lines joining P to the two
1
If P(x, y) is any point on the locus,
complex plane, verifies |z − 3| + |z + 3| = 10
5. (a) Determine the Cartesian equation of the locus of a point P(x, y) whose affix z, in the

d(P, F) = d(P, l)
then Squaring both sides:
(x − 4)2 + y2 =
1 (x 16)2
(b) Find the characteristic elements of the locus
d(P, F) = (x − 4)2 + (y − 0)2 x2 − 8x + 16 + y2 =
2 4
x2 32x 256 (center, lengths of semi-axes, foci, directrices and vertices)

4x2 + 4y2 − 32x + 64 = x2 − 32x + 256

4 (c) Plot its curve on the Cartesian plane.

= (x − 4) + y2 2
3x2 + 4y2 = 192
Ex 1: Write the equation of the tangents line to the ellipse x2 + 4y2 − 20 = 0, one at the
5.2.3 Tangents to an ellipse
Divide both sides by 192:
d(P, l) = = |x − 6| + =1
|x 6| x2 y2 point whose abscissa is 4 and its ordinate is positive, and the other at the point
∴ the locus is an ellipse.
12 02 4864
whose ordinate is 2 and its abscissa is positive. Verify that these tangents are
concurrent at the point A(10/3, 5/3).

Sln: x2 + 4y2 − 20 = 0 ⟹ 2x + 8yy' = 0 m2 = f' (2, 2) = − 4(2) = − 4

Ex5: Find the equation of the ellipse having its center at the origin, major axis on the x-axis and

y' = − 4y
2 1

∴ T2 ≡ y − 2 = − (x − 2)
passing through the points (4, 3) and (6, 2). x
1

Substitute a2 = 52in (1): by T ≡ y − y0 = m(x − x0 ) where x + 4y = 10.

− − = −4
Sln: The standard form is:
+ 2=1
The tangent at any point of the ellipse is given 4
64 36

+ 2=1
x2 y2

(1) Substitute x = 4, y = 3: + m = f' (x0 , y0 ) (1) x + y = 5 −1 −x − y = −5

a2 b2 Intersection of T1 and T2
 +
a2 b 16 9

+ 2=1 + =9 = 1 − 13 At the first point: x0 = 4, (2) x + 4y = 10 1 x + 4y = 10

52 b

42 + 4y20 − 20 = 0
16 9 324 36 9 4

(2) Substitute x = 6, y = 2: = 13 3y = 5
=5
2 2
a2 a2 b b

4y20 = 4
b

y = 5/3
9 9

+ 2=1
260

a = = 13 y20 = 1 Substitute y = 5/3 in (1):

36 4 a2 b2
260 1 1

b = 13 y0 = 1 (since y0 ≥ 0) x+3=5
2

(1) ' 2 + 2 = 1
a2

−4
b
b2
16 9 5 5

( a2 = 52
2

m1 = f' (4, 1) = − = −1 x=5−3

(2) & 2 + 2 = 1
a b So the required equation is:
+ 13 = 1
4 5

%a ∴ T1 ≡ y − 1 = −1(x − 4) x= 3
36 4 x2 y2 4(1)
9
x + y = 5.
10

∴ T1 ∩ T2 = -(10/3, 5/3. as required

b 52

At the 2nd point: y0 = 2,

x0 2 + 4*22 + − 20 = 0
x0 2 = 4
x0 = 2 (since x0 ≥ 0)

Prepared by Jean Paul RUBAYIZA Page 241 Prepared by Jean Paul RUBAYIZA Page 242
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 x  2cosθ 1. Find the equation of the tangent and normal line to the ellipse x2 + 2y2 = 9 at
Ex3: Find the equation of the tangent to the ellipse Exercise 5.5
 at the point where θ  6.
y  sinθ the point (1, − 2).
π

2. Find the values of c if y = 3x + c is tangent to the ellipse x2 + 4y2 = 4.

x  2cos  √3 where θ = :
Sln: Coordinates of the point of tangency P Slope of the tangent line at the point

∴ P(√3, )
π


π

y  sin 6  2
√3
6 1 6 Find the coordinates of the point of contact.
m = dx = 21 = − 2
√3
π 1

x'(θ) = −2sinθ
2

x = 5cosθ, y = 3sinθ at the point where θ = − .

dy 3. Find the equation of the tangent and the normal to ellipse


y'(θ) = cosθ
2π
1

T ≡ y − y0 = m(x − x0 ) 4. Find the equations of the tangent lines with slope m = 2 to the ellipse + = 1.
The tangent line at P(√3, ) is given by 3

x'( ) = −2sin = −1
2
x2 y2

For θ = 6, /
π π

y − = − (x − √3)
√3
4 9

y'( ) = cos =
√3
π 6 6 1
5. Show that the equation of the tangent line at (x0 , y0 )to the ellipse
2y − 1 = −√3x + 3
π π

+ = 1 is + =1
2 2

√3x + 2y = 4
6 6 2 x2 y2 xx0 yy0
b2 b2

Ex 2: Discuss according to the values of the parameter m, the position of the line l ≡ x + y = m
a2 a2

and the ellipse E ≡ 4x2 + y2 = 1? Exercise 5.6

(1) x + y = m ∆= ( − 2m)2 − 4 × 5(m2 − 1)
 25x2 + 4y2 + 50x − 16y = 59
1. Find the foci, the vertices and sketch the graph of
(2) 4x2 + y2 = 1 = 4m2 − 20m2 + 20
Sln:
From (1): y = m − x = −16m2 + 20
2. (a) Find the equation of the ellipse having its center at (1, 2), focus at (6, 2) and
Replace y by m − x in (2):
containing the point (4, 6).

4x2 + (m − x)2 = 1 Sign of ∆ lines joining P(x, y) to (3, − 2) and ( − 2, 1) is −6.

(b) Determine the locus of a point P(x, y) so that the product of the slopes of the

4 + (m2 − 2mx + x2 ) = 1 ∆= 0 ⟺ −16m2 + 20 3. For what value of p the line y = 2x − p is tangent to the ellipse 9x2 + 16y2 = 144?
4x2 + m2 − 2mx + x2 − 1 = 0 m2 = 4. Find the equation of the ellipse with a focus at (1, 0), directrix x + y + 1 = 0 and
5x2 − 2mx + m2 − 1 = 0
5

the eccentricity e = 2
√2
m=±
√5
4

from the line x + 2 = 0. Derive the equation of its locus.

2
Synthetic table of discussion 5. A point P(x, y) moves so that its distance from the point (3, 2) is one half of its distance

∆= −16m2 + 20
m Sign of Discussion What curve is it?

−∞
6. Derive the equation of the ellipse having its center at the origin, one focus at

(a) Find the equation of the tangent and normal to the ellipse 2x2 + 3y2 − 30 = 0
(0, 3) and the length of semi – major axis 5.

− at the point ( − 3, 2).

7.

√5 (b) Find the equations of the tangents to the ellipse 5x2 + 7y2 = 35 which are
Line l is outside of the ellipse E

− 0 Line l is tangent to the ellipse E perpendicular to the line 3x + 3y − 12 = 0.

2
+ 3*z2 + z̅2 + + 10zz̅ = 4?
8. (a) In the complex plane, what is the set E of all points P whose affix z verifies

√5
Line l is secant to the ellipse E
(b) Determine the coordinates of the foci.
0 Line l is tangent to the ellipse E (c) Determine the coordinates of the most points.
2
−
(d) Find the eccentricity.
Line l is outside of the ellipse E
+∞
(e) Write the equations of directrices of the set E.
(f) Represent graphically the set E.

Prepared by Jean Paul RUBAYIZA Page 243 Prepared by Jean Paul RUBAYIZA Page 244
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 9. (a) Lines drawn from the foci to any point on an ellipse are called the focal radii 5.3.2 Cartesian and parametric equations of a hyperbola

the ellipse 3x2  4y2  48. Illustrate your results on a graph. F2 P  F1 P  2a

of the ellipse. Find the equations of the focal radii drawn to the point (2, 3) on Let P(x, y) be any point on the hyperbola, then from the definition
(1)

the ellipse x2  5y2  2x  5y  20  0. F1 P  (x  c)2  (y  0)2  (x  c)2  y2

(b) Find the equations of the focal radii drawn to the point (2, 3) on

10. (a) Show that the curve defined by the equation y  7  5 √16  6x  x2
2

(x 3)2

(y 7)2
 1, F2 P  (x  c)2  (y  0)2  (x  c)2  y2
∴ Equation (1) becomes:
is a semi – ellipse of the equation
which lies above the line y  7.
25 4

(x + c)2 + y2 − (x − c)2 + y2 = 2a

(b) Sketch the given curve in part (a.)

(x + c)2 +y2 = 2a + (x − c)2 +y2

5.3. The hyperbola

5.3.1 Definition: foci and directrices of a hyperbola Squaring: (x + c)2 + y2 = 4a2 + (x − c)2 + y2 + 4a(x − c)2 + y2

x2 + 2cx + c2 + y2 = 4a2 + x − 2cx + c2 + y2 + 4a(x − c)2 + y2

2

4cx − 4a2 = 4a(x − c)2 + y2

cx − a2 = a(x − c)2 + y2
The path of a point which moves so that the

F1 (c, 0) and F2 (  c, 0) is 2a, where a is a

difference of its distances from two fixed points

constant and a  c, is a hyperbola. Squaring again: c2 x2 − 2ca2 x + a4 = a2 (x − c)2 + y2 !

c2 x2 − 2ca2 x + a4 = a2 "x2 − 2cx + c2 + y2 #
c2 x2 − 2ca2 x + a4 = a2 x − 2ca2 x + a2 c2 + a2 y2
The two fixed points F1 (c, 0) and F2 (  c, 0)
2

a x − c2 x2 + a2 y2 = a4 − a2 c2
• 2 2

(a2 − c2 )x2 + a2 y2 = a2 (a2 − c2 )

are foci of the hyperbola.

Multiplying by −1, (c2 − a2 )x2 − a2 y2 = a2 (c2 − a2 )

• The equations of the directrices d1 and d2 are

d1 ≡ x  e and d2 ≡ x   e . Dividing each term by a2 (c2 − a2 ), the equation becomes

given by
a a

− c2 =1
x2 y2

As a < c, then c − a > 0. Let c2 − a2 = b2 , then we have the standard form of the
a2 a2
2 2

equation of a hyperbola,

− = 1 or b2x2 − a2 y2 = a2 b2
x2 y2
a2 b2

If the foci F1 and F2 had been at (0, c) and (0, − c) respectively, the standard form would be:

− =1
y2 x2
a2 b2

Prepared by Jean Paul RUBAYIZA Page 245 Prepared by Jean Paul RUBAYIZA Page 246
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 If the center was at the point (h, k) the equation would be

 1
(x h)2 (y k)2
a2 b2

This equation can be represented parametrically by:

x  h  asecθ

y  k  btanθ Curve

 secθ  sec2 θ
(x h)2

⟹ 
x h

 tanθ 1  tan2 θ  sec2 θ

a2 Recall that:
 tan2 θ
a
In fact: y k (y k)2

sec2 θ  tan2 θ  1
b b2

 1
(x h)2 (y k)2
a2 b2
In particular, the parametric equations of the hyperbola centered at the origin are
x  asecθ

y  btanθ
Ex1: Given the hyperbola 9x2  16y2  144,
  1 is expanded, it takes the form:
(x h)2 (y k)2
When the equation (a) Find the coordinates of its vertices. (b) Find the coordinates of its foci.
a2 b2

Ax2  By2  Cx  Dy  F  0
(c) Find the equations of asymptotes. (d) Find the eccentricity and the equations
(e) Plot the graph of the hyperbola. of directrices.

Sln: (a) 9x2  16y2  144,

where A and B have the same sign
(e) The length of the latus rectum:
 1 LR    4.5
x2 y2 2b2 29
5.3.3 Characteristic elements of the hyperbola Divide by 144:
16 9 a 4

 1  1
2 2 2 2 Center is at the origin,
Equation (x h) (y k) (x h) (y k)

a2  16, b2  9
transverse axis is the x-axis
a2 b2 b2 a2

Focal length c a2  b2 a  4, b  3
Vertices: V1, 2  (  a, 0)
e  ∴ V1  (4, 0), V2  (  4, 0)
Eccentricity c a2 b2

(b) Foci: F1, 2  (  c, 0)

a a

LR 
c  a2  b2  √16  9
Length of latus 2b2

V1, 2  (h  a, k) V1, 2  (h, k  a)  √25  5

rectum a

∴ F1  (5, 0), F2  (  5, 0)
Vertices
x  h, y  k x  h, y  k
(c) Asymptotes: 16  0
Axes of symmetry
x2 y2

F1, 2  (h  c, k) F1, 2  (h, k  c)

y  4x
9
Foci 3

y  k   (x  h) y  k   (x  h) (d) Eccentricity: e  
Asymptotes b a c 5

Directrices: d1, 2 ≡ x   e
b a 4
a
d1, 2 ≡ x  h  d1, 2 ≡ y  k  e
a

4 
a a
Directrices a 4 16
e

∴ d1, 2 ≡ x  
e 5 5
16
5

Prepared by Jean Paul RUBAYIZA Page 247 Prepared by Jean Paul RUBAYIZA Page 248
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 Ex4: Show that the curve with parametric equations x  4  2tant, y  1  3sect is a hyperbola.
9x2  16y2  18x  64y  199  0
Ex2: Consider the following hyperbola defined by
State the coordinates of the center, lengths of the semi-axes, vertices, foci, equations of
(a) Write its equation in the standard form asymptotes and sketch the curve.
(b) Hence find:

x  4  2tant tant  2 tan2 t 

(i) the center (ii) the vertices (iii) the equations of asymptotes. (x 4)2

Sln: :  ⟺ / y 1 ⟹ /
x 4
(1)
y  1  3sect 1  tan2 t  sec2 t
(c) Sketch the graph of the given hyperbola. Recall that:
sect  sec2 t 
4

Sln: (a) 9x2  16y2  18x  64y  199  0 sec2 t  tan2 t  1

(y 1)2
3 (2)
9

(2)  (1):   sec2 t  tan2 t

(y 1)2 (x 4)2

9x2  2x  16y2  4y  199  0

Completing the squares: (c) The length of the latus rectum:
LR    4.5
2b2 9 4

9x2  2x  1  1  16y2  4y  4  4  199  0

29

9x2  2x  1  16y2  4y  4  199  9  64  0  1

a 4
(y 1)2 (x 4)2
which is the Cartesian equation of a hyperbola.
9(x  1)2  16(y  2)2  144
9 4

 9 1
(x 1)2 (y 2)2

h  1, k  2 h  1, k  2
16
Transverse axis // to the y-axis:

a2  16, b2  9
C  (h, k)  (1,  2)
Coordinates of the center:
a  4, b  3
(b) (i) Center: C  (h, k)  (1,  2)
(ii) Vertices: V1, 2  (h  a, k)  (1  4,  2)
a2  9, b2  4
Lengths of semi-axes:
∴ V1  (5,  2), V2  (  3,  2)
a  3, b  2
 0 Vertices: V1, 2  (h, k  a)  (4, 1  3)
(x 1)2 (y 2)2
(iii) Asymptotes:
y  2   4 (x  1) ∴ V1  (4, 4), V2  (4,  2)
16 9
3

(iv) Foci: F1, 2  (h  c, k)

Foci: F1, 2  (h, k  c)
c  a2  b2  √16  9  5
∴ F1, 2  (1  5,  4) c  a2  b2  √9  4  √13
F1  (6,  2), F2  (  4,  2) ∴ F1, 2  (4, 1  √13)
F1  (4, 1  √13)  (4, 4.6)
F2  (4, 1  √13)  (4,  2.6)

line 4y  9  0. Find the locus of the point P.

Ex3: A point P(x, y) moves so that its distance from the point A(0, 4) is 4/3 its distance from the
 0
Sln: A(0, 4), l ≡ 4y  9  0
(y 1)2 (x 4)2
Asymptotes:
y  1   2 (x  4)
Squaring both sides: 9 4

x2  (y  4)2 
3
If P(x, y) is any point on the locus, (4y 9)2

d(P, A)  3 d(P, l)
x2  y2  8y  16 
4 9
then 16y2 72y 81 Ex5: Write the equation of the hyperbola passing through the point (4, 6) and whose
d(P, F)  (x  0)  (y  4)  x2  (y  4) 9x2  9y2  72y  144  16y2  72y  81 asymptotes are y  √3x.
2 2 2 9

|4y 9| |4y 9| 9x  7y  63, Sln: y  √3x ⟺ *y  √3x+*y  √3x+  0 36  3  16  c

d(P, l)  2 2 
Divide both sides by 63:
2 2

c  12
∴ the required equation is
4

 7 1
0 4

∴ x2  (y  4)2   *y  √3x+*y  √3x+  c

4 |4y 9|

The required equation has the form:
y2 x2

∴ the locus is an ellipse. y2  3x2  c y2  3x2  12

9

3x2  y2  12
3 4

x2  (y  4)2  Substitute x  4, y  6:

|4y 9|

62  3(4)2  c
3

Prepared by Jean Paul RUBAYIZA Page 249 Prepared by Jean Paul RUBAYIZA Page 250
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex6: Determine the equation of the hyperbola with its center at (  4, 1), vertex at (2, 1) and 5.3.4 Equilateral and conjugate hyperbolas
length of semi-conjugate axis 4.

Sln: C(  4, 1), V(2, 1)

  1:
(x h)2 (y k)2
Substitute these values in
h  4, k  1
b2

 1
Transverse axis // y-axis a2 (1) Equilateral hyperbola
(x 4)2 (y 1)2

Semi-conjugate axis: b  4
Semi-transverse axis: a  d(C, V)
If a  b, the equation a2   1 becomes:
36 16
x2 y2
b2

a  (2  4)2  (1  1)2  62  02  6 x2


y2
 1 ⟺ x2  y2  a 2
a2 a2

Ex7: Determine the equation of the hyperbola having its center at the origin, transverse axis on
Its asymptotes are y  x and they are perpendicular one each other.
This equation represents an equilateral (or rectangular) hyperbola.
the x-axis, the eccentricity √7, and the length of the latus rectum 6.
1

The transverse and conjugate axes are equal and so its eccentricity is e  √2.
 1
2
x2 y2
Sln: The required equation is in the form
a2 b2

Ex: The equation of a curve C is y  √x2  4.

(1) e  a  
√7
(1) 3a2  4b2  0 Substitute a  4 in (2)
c a2 b2


a2  b2  b2  3a b2  3(4)
(a) Show that the curve C is a semi – hyperbola of the rectangular hyperbola
x2  y2  4, which lies above the x-axis ( y  0)
a 2
a√7 (2)
b2  12
a2  b2 
2

Substitute b2  3a in (1):
7a2 (b) Sketch the curve C.
4a  4b  7a 3a2  4(3a)  0
So the required equation is:
 12  1
4
2

3a(  a  4)  0
2 2

3a2  4b2  0 Sln: C ≡ y  √x2  4 Asymptotes: x2  y2  0

x2 y2

a  0 to be rejected y  x
16

a  4 and a2  16 y2  x2  4
Squaring both sides:
(2) LR  6
2b2

x2  y2  4
2b  6a
a

x2  y2  4 (a rectangular hyperbola)
2

b2  3a
∴ y  √x2  4 is represented by the
positive branches of the hyperbola
  1 because √t ) 0 where t ∈ 
x2 y2

1. Show that the curve with parametric equations x  5tant, y  3sect is a hyperbola.
Exercise 5.7
a2  b2  4
4 4

ab2
State the coordinates of the center, lengths of the semi-axes, vertices, foci,
Vertices: V1, 2  a, 0  2, 0
2. For the hyperbola 9x2  16y2  36x  32y  124  0, find
equations of asymptotes and sketch the curve.
Foci: F1, 2  c, 0  2, 0
c  a2  b2  √4  4  √8
(a) the coordinates of the center (b) the foci (c) the vertices

 2√2
(d) the equations of asymptotes (e) Sketch the hyperbola.

∴ F1, 2  (  2√2, 0)  (  2.8, 0)

is 3/2 of its distance from the line y  8/3  0.
3. Find the equation of a locus of a point P(x, y) which moves so that its distance from (0, 6)

and passing through the points (4, 6) and (1,  3).

4. Find the equation of the hyperbola with center at the origin, transverse axis on the y-axis,

5. A point moves so that the difference of its distances from (0, 3) and (0,  3) is 5. Derive
the equation of its locus.

Prepared by Jean Paul RUBAYIZA Page 251 Prepared by Jean Paul RUBAYIZA Page 252
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 (2) Conjugate hyperbola hyperbolas 5.3.4 Tangents and normal lines at a hyperbola

3x2  y2  3 at the point (2, 3).

Two hyperbolas are called conjugate hyperbolas if the transverse and conjugate Ex1: Find the equation of the tangent and normal line to the hyperbola
axes of one are respectively the conjugate and transverse of the other.
Thus a2  b2  1 and  a2  b2  1 are conjugate hyperbolas.
x2 y2 x2 y2

Sln: Substitute x  2, y  3:
  1. 3(2)2  (3)2  12  9  3 T ≡ y − y0 = m(x − x0 )
x2 y2
Tangent at (2, 3) is given by:

∴ Point (2, 3) lies on the hyperbola y − 3 = 2(x − 2)

Ex: Write the equation of the hyperbola conjugate to the hyperbola
9 16

3x2 − y2 = 3. y = 2x − 1
Sketch the two hyperbolas on the same graph.

3x2 − y2 = 3 ⟹ 6x − 2yy' = 0
1st Hyperbola
N ≡ y − y0 = − m (x − x0 )
Sln Conjugate hyperbola Slope of the tangent at any point Normal line at (2, 3) is given by:
1

 1   1 y' = y y − 3 = − (x − 2)
2 2
Equation x y x2 y2 3x
1

a3 a4 2y − 6 = −x + 2
9 16 9 16 2

m = f' (2, 3) = 3 = 2 x + 2y = 8
Semi-transverse axis Slope of the tangent line at (2, 3):
b4 b3
3(2)
Semi- conjugate axis
V1, 2  (  a, 0)  (  3, 0) V3, 4  (0,  a)  (0,  4)
Ex2: Find the angles of intersection of the curves 2x2 + y2 = 20 and 4y2 − x2 = 8
Vertices
Focal length c a2  b  √9  16  √25  5
2 Sln: The angle of intersection is defined to be the angle between their tangents at their
Foci F1, 2  (  c, 0)  (  5, 0) F3, 4  (0,  c)  (0,  5) point of intersection.

2b2 2  16 2b2 2  9
LR    10.5 LR    4.5 (1) 2x2 + y2 = 20 C1 ≡ 2x2 + y2 = 20 ⟹ 4x + 2yy' = 0
Length of latus
3
(2) −x2 + 4y2 = 8 y' = −
rectum a 3 a 4 1
 16  0 ⟺ y  3x
2x
Asymptotes x2 y2 4 2
C2 ≡ 4y2 − x2 = 8 ⟹ 8yy' − 2x = 0
y

2x + y = 20
9

+ y' =
−2x2 + 8y2 = 16
2 2 x
4y

8y = 36
y2 = 4
2
At the point (2√2, 2):

y = ±2 m1 = − = −2√2
2(2√2)

Substitute y2 = 4 in (1) m2 = 4(2) =

√2
2

2x2 + 4 = 20
2√2

2x2 = 16
4
√2
tanθ = =( (= =∞
7√2

x2 = 8
m1 m2 2√2
√2
2√24 5
4 4

x = ±2√2 θ = 90° i.e. the two curves are orthogonal.

Curve sketching 1 m1 m2 1 0
4

( ± 2√2, 2) and ( ± 2√2, − 2)

The points of intersection are:
By symmetry, the curves are orthogonal at each of
their points of intersection

Prepared by Jean Paul RUBAYIZA Page 253 Prepared by Jean Paul RUBAYIZA Page 254
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
Ex3: If the line y  2x  1 is tangent to the hyperbola x2  ky2  1,
1. The equation of a curve C is −9x2 + 16y2 + 36x + 32y − 164 = 0.
Exercise 5.8
find the value of k.

(1) y = 2x + 1 ∆= 16k2 − 4(4k2 − 4k − k − 1)

(a) Determine the standard form of the curve C. Which shape is it?

(2) x2 − ky2 = 1 = 16k2 − 4(4k2 − 5k − 1)
(b) Find the coordinates of the center, vertices, focal points and the equation
Sln:
Replace y by 2x + 1 in (2): = 16k2 − 16k2 + 20k + 4
of the focal line.

x2 − k(2x + 1)2 = 1 = 20k + 4

(c) Write the equations of asymptotes to the curve C.

x − k(4x2 + 4x + 1 ) = 1
2. Determine the nature of the curve defined by y = −1+ 3 √x2 − 4x − 5 and
(d) Sketch the curve C.

x2 − 4kx2 − 4kx − k − 1 = 0 ∆= 0
2
The point of tangency occurs when 2

4kx2 − x2 + 4kx + k + 1 = 0 20k + 4 = 0

(4k − 1)x2 + 4kx + k + 1 = 0
sketch its graph.
k = −5
∆= (4k)2 − 4(4k − 1)(k + 1) asymptotes are y = ±√2x.
1 3. (a) Write the equation of the hyperbola passing through the point (1, 2) and whose

(6, 0) and the equation of one asymptote 4x − 3y = 0.

(b) Find the equation of the hyperbola having its center at the origin, one vertex at
c c

4. Find the equation of the hyperbola whose directrix is the line 2x + y = 1, whose one
Ex4: Tangents at the point R(cp, p ) and S(cq, q ) to the rectangular hyperbola
xy  c2 intersect at the point T.
(a) Show that the coordinates of point T are x 
2cpq
and y 
2c
. focus is at (1, 2) and whose eccentricity is e = √3.

(a) Transverse axis 8, foci at ( ± 5, 0)

p q p q 5. Find the equations of the hyperbolas for which the following conditions are given

the condition pq  2.
(b) Determine the locus of the point T, such that p and q vary according to
(b) Conjugate axis 24, foci at (0, ± 13)
Sln: (a) xy  c2 ⟹ y  xy'  0 (b) x =
(c) Center at (0, 0), a focus at (8, 0), a vertex at (6, 0).
y'   x (1) x + p2 y = 2cp
Finding the coordinates of T 2cpq
(1)
3 y= the slopes of the lines joining it to ( − 2, 1) and (4, 5) is 3.
y

(2) x + q2 y = 2cq −1
1
p q 6. Derive the equation of the locus of a point P(x, y) which moves so that the product of
2c
c (2)
At the point R(cp, p ): p q

x + p2 y = 2cp
2c 7. Determine the equation of a hyperbola with its axes parallel to the coordinate axes
m1   cp   p  cp   p2
Replace by y in (1):
+
c

x=
p q

−x − q2 y = −2cq
p c 1 1 and center at the origin, if the latus rectum is 18 and the distance between the
T1 ≡ y − y1 = m1 (x − x1 )
2c

x = ypq
pq
8. Determine the equation of the tangent to the hyperbola xy = 12 at the point P with
foci is 12.
(p − q y = 2cp − 2cq
p q

∴ T1 ≡ y − = − (x − cp) Substitute pq = 2 in (3): abscissa x = 6. This tangent cuts the coordinate axes respectively at points R
c 1 2 2 (3)

p2y − pc = −x + cp x = 2y and Q. Show that PQ = PR.

y= = (p =p
p p2

x + p2 y = 2cp y=2
9. (a) Find the equation of the tangent and the normal to the hyperbola x2 − 4y2 = 4
2cp 2cq 2c(p q) 2c
x

Substitute y = ∴ the locus of the point t when

p2 q2 q)(p q) q

at the point ( − 2√5, 2).

2c

p and q are related by pq = 2

c
At the point S(cq q ): in (1):
x + p2 p q = 2cp
p q

m2 = − cq = − q × cq = − q2 is the line y =
c 2c
q c 1 1 x (b) Find the Cartesian equation of the tangent line to each of the following curve
T2 ≡ y − y2 = m2 (x − x2 ) x = 2cp −
x = 3secθ
2cp2 2
at the specified point.

∴ T2 ≡ y − = − (x − cq)  , θ=4
x = 2cp 1 − p q y = 5tanθ
p q π
c 1 p ,
q2 y − qc = −x + cq 10. If the line y = px + 5 is tangent to the hyperbola 9x2 − 16y2 = 144,
x = 2cp p q =
q q2

x + q2 y = 2cq
q 2cpq
p q find the value of p.

∴ x= and y = p
2cpq 2c
p q q

Prepared by Jean Paul RUBAYIZA Page 255 Prepared by Jean Paul RUBAYIZA Page 256
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
5.4. General equation of Conics
(2) Classification of conics

Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0

5.4.1 Reduction of the general equation of a conic Consider the general equation of a conic

f(x, y) ≡ Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0

The general equation of a conics is
(1) Its characteristic binomial (δ) and its discriminant (∆) are respectively

δ= , ∆ = (B C E (
where at least one of the coefficients A, C or F is not zero A B D
A B

(x, y  A B  y  + 2(D, E  y  + F = 0
In matrix form, equation (1) can be written as:

• If A = C and B = 0, the conic is a circle.

x x B C
(2) D E F

• If δ = AC − B > 0, the conic may be an ellipse.

B C

• If δ = AC − B2 = 0, the conic may be a parabola.

or 2

( x, y, 1 7B C E 8 9 y : = 0
A B D x
• If δ = AC − B2 < 0, the conic may be a hyperbola.
(3)
D E F 1
• For special cases which may exists, the conic will be two straight lines,
The determinant δ =
A B a point, or imaginary.
• is called characteristic binomial of the conic.
B C
The determinant ∆ = (B C E ( is called the discriminant of the conic.
A B D NATURE
∆≠0 ∆=0
•

δ>0
D E F TYPE

Ex1: Show that the point (2, − 1) lies on the conic (∆ < 0)
Ellipse Non-degenerated ellipse Point-ellipse

x2 + 8xy + 3y2 + 8x − 10y − 17 = 0 Imaginary ellipse (∆ > 0)

(NDE)

δ<0
22 + 8(2)( − 1) + 3( − 1)2 + 8(2) − 10( − 1) − 17 = 2 − 16 + 3 + 16 + 10 − 17 = 0.
Sln:
Hyperbola Non-degenerated hyperbola Degenerated hyperbola in
∴ (2, − 1) lies exactly on the conic x2 + 8xy + 3y2 + 8x − 10y − 17 = 0. δ=0
(NDH) two intersecting lines.
Parabola Non-degenerated parabola Degenerated parabola in
(NDP) two parallel lines.
Ex2: Write down a 3 × 3 matrix representing the conic
x2 − 2√3xy + 3y2 + 16√3x + 36 = 0
(a) 3x2 + 3y2 − 12x + 12y − 1 = 0
Ex1: Determine the type and the nature of each of the following conics:

A = 1, 2B = −2√3 C = 3, 2D = 16√3 2E = 0 F = 36 (b) 2x2 − 2xy + y2 + x − 2y + 1 = 0

Sln:

B = −√3 D = 8√3 E=0 (c) x2 + 5y2 + 2x − 20y + 25 = 0

(d) x2 + 4xy + 4y2 − 4y + 3 = 0
Sln: (a) 3x2 + 3y2 − 12x + 12y − 1 = 0
1 −√3 8√3 A = C = 3 and B = 0, so the conic is a circle.
The matrix representation of the given conic is

M = ;−√3 0 ; (b) 2x2 − 2xy + y2 + x − 2y + 1 = 0

A = 2, 2B = −2, C = 1, 2D = 1, 2E = −2, F=1
3

B = −1 D = 0.5 E = −1
8√3 0 36
2 −1
δ= = = 2 − 1 = 1,
−1
A B

2 −1 0.5
B C 1
∆ = ( B C E ( = ( −1 1 −1 ( = −0.25
A B D

0.5 −1
δ=1>0
= ⟹ the conic is an ellipse.
D E F 1

∆ = −0.25 < 0

Prepared by Jean Paul RUBAYIZA Page 257 Prepared by Jean Paul RUBAYIZA Page 258
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 (c) x2  5y2  2x  20y  25  0
A  1, 2B  0, C  5, 2D  2, 2E  20, F  25
Discussion

B0 D1 E  10 ∞

λ δ ∆ Conic

δ   5  0  5,
  
A B 1 0
B C 0 5 Imaginary ellipse
∆  ( B C E (  (0 5 10(  30
A B D 1 0 1
1 
1 10
0 Parabola
D E F 25

δ10   
= ⟹ the equation represents an imaginary ellipse.
Hyperbola
∆  30  0

(d) x2  4xy  4y2  4y  3  0
0 0 Two intersecting lines (DH)

A  1, 2B  4, C  4, 2D  0, 2E  4, F  3    Hyperbola
B2 D0 E  2

δ   44  0
A B 1 2 1 0 Parabola

  
B C 2 4
∆  ( B C E (  (2 4 2(  4
A B D 1 2 0 Ellipse

0 2 ∞
δ0
= ⟹ the conic is a parabola.
D E F 1

∆  4 ≠ 0
Ex3: Show that the equation 2x2  5xy  12y2  0 is a union of two intersecting lines and find

λx2  2xy  λy2  2(λ  1)x  2y  2  0 A  2, 2B  5, C  12, 2D  0, 2E  0, F  0

Ex2: Discuss, according to the values of the parameter λ, the type and the nature of the conic: the acute angle between them.

A  λ, 2B  2, C  λ, 2D  2(λ  1), 2E  2, F  2 B  2.5 D0 E0

Sln:

B  1 D  λ  1 E1 δ   24  6.25  30.25

λ 1 2.5 12
A B 2 2.5
δ   λ2  1
1
A B B C
∆  ( B C E (  (2.5 12 0(  0
A B D 2 2.5 0
1 λ  1
B C λ
∆  ( B C E (  ( 1 1 (  λ
A B D λ
δ  30.25  0
3

= ⟹ the equation represents a hyperbola degenerated in two

D E F 0 0 0
λ  1
λ
∆0≠0
D E F 1 2

Taking 2x2  5xy  12y2 as a polynomial in x:

intersecting lines.
δ  0 ⟺ λ2  1  0 ⟺ λ  1
Sign of δ
Delta: ∆  (5y)2  4(2)(12y2 )  25y2  96y2  121y2, √∆  11y
∞ 1 ∞
δ  λ2  1    x1  4  4  2
λ 1 5y 11y 6y 3y

x2    4y
0 0
5y 11y 16y

2x  5xy  12y  0 ⟺ (2x  3y)(x  4y)  0

4 4

∆  0 ⟺ λ3  0 ⟺ λ  0
Sign of ∆
∴ 2x2 + 5xy − 12y2 = 0 is the combined equations of the two lines:
2 2

∞ ∞ l1 ≡ 2x − 3y = 0 and l2 ≡ x + 4y = 0
∆  λ3  
λ 0
m1 = 3 m2 = − 4
0 2 1

tanθ = =  =  12
10  = = 1.1
2 1 11
m1 m2 11
 
3 4

θ = arctan1.1 = 47.7° ∴ the angle between the two lines is 47.7°

2 1
1 m1 m2 1 10
3 4 12

Prepared by Jean Paul RUBAYIZA Page 259 Prepared by Jean Paul RUBAYIZA Page 260
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 Exercise 5.9 Ex1: Given the equation of the ellipse E ≡ 9x2  25y2  72x  150y  144  0

λx2  2xy  2λx  8  0

1. For what value of λ the point P(1, 3) lies on the conic: (a) By completing squares, write the equation of the ellipse (E) in standard form.
(b) Find the coordinates of the center, semi-axes, vertices and the foci of the ellipse (E).

(a) 3x2  10xy  3y2  x  32  0 (d) xy  x  2y  3  0

2. Determine the type and the nature of each of the following conics (c) Sketch the curve showing all two sets of axes.

(b) 41x2  84xy  76y2  168 (e) x2  4xy  4y2  4 (a) 9x2  25y2  72x  150y  144  0
(c) 16x  24xy  9y  30x  40y  0
Sln:

9x2  8x  25y2  6y  144  0

2 2 Completing the squares:

ax2  2axy  y2  2x  2ay  a  0 (a ∈ ℝ) 9x2  8x  16  16  25y2  6y  9  9  144  0

3. Discuss, according to the values of the parameter a, the type and the nature of the conic:

9x2  8x  16  25y2  6y  9  144  144  225  0

5.4.3 Reduction of conics 9(x  4)2  25(y  3)2  225
 9 1
Guidelines: (x 4)2 (y 3)2

Letting x  4  x'and y  3  y', the equation becomes:

Step1: Find the center of the conic C(h, k) 25

f 'x = 2Ax + 2By + 2E = 0 Ax + By + E = 0

(h, k) is a solution of the system
3 ⟺   9 1
f 'y = 2Bx + 2Cy + 2D = 0
x'2 y'2

Bx + 2Cy + D = 0 (b) By identification h  4, k  3 and so the center is at C(4, 3)

25

Step2: Perform a translation of the origin of axes to the center of the conic. a2  36, b2  16
a  5, b3
Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0
After translation the equation of the conic

V1, 2  (  a, 0)  (  5, 0)
Vertices in new coordinates:

Ax'2 + 2Bx'y' + Cy'2 + F' = 0

becomes:
V3, 4  (0,  b)  (0,  3)
x' = x − h
where 
y' = y − k
(Formulas of translation) and Foci in new coordinates:
F' = f(h, k) c  a2  b2  √25  9  √16  4
Step3: Find the angle θ through which it is necessary to rotate the axes in order to ∴ F1, 2  (  c, 0)  (  4, 0)
(c) The length of the latus rectum: LR     3.8
2b2

tan2> =
eliminate the xy-term by solving equation 29 18
2B a 5 5
A C

A'x''2 + C'y''2 + F'' = 0

After rotation of axes, the equation of the conic becomes:

x' = x''cos> − y''sin> cos> −sin> x''

where  ⟺ 4 5= 4 5
y' = x''sin> + y''cos>
x'
y' sin> cos> y''
(formulas of rotation)

Prepared by Jean Paul RUBAYIZA Page 261 Prepared by Jean Paul RUBAYIZA Page 262
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 8x''2 + 2y''2 = 8
5x2  6xy  5y2  4x  4y  4  0. 4x''2 + y''2 = 4
Ex2: By translation and rotation of axes, reduce the equation

A  5, 2B  6, C  5, 2D  4, 2E  4, F  4 x''2 + =1
Sketch the curve showing all three sets of axes. y''2
(Ellipse)
B3 D  2 E2
Sln: 4

Center of the conic: f(x, y)  5x2  6xy  5y2  4x  4y  4  0

f 'x  10x  6y  4  0 (1) 5x  3y  2
(i)
3 ⟺  ? a = 2, b=1
Lengths of semi-axes:
f 'y  6x  10y  4  0 (2) 3x  5y  2 3
5

V1, 2 = (0, ± a) = (0, ± 2)

Vertices in new coordinates:
25x  15y  10 Substitute x  1 in (1):

9x  15y  6 V3, 4 = ( ± b, 0) = ( ± 1, 0)
5  3y  2
3y  3 Foci in new coordinates: F1, 2 = (0, ± c)
16x  16
x1 y  1 c = a2 − b2 = √4 − 1 = √3
∴ The center is at C(1, − 1); h = 1 , k = −1 ∴ F1, 2 = (0, ± √3) = (0, ± 1.7)
The length of the latus rectum:
LR = = =1
Ax'2 + 2Bx'y' + Cy'2 + F' = 0
After the translation of axes, equation (i) becomes: 2b2 2×1

x' = x − 1
a 2

where 
y' = y + 1
f(x, y) = 5x2 + 6xy + 5y2 − 4x + 4y − 4
F' = f(1, − 1) = 5(1)2 + 6(1)( − 1) + 5( − 1)2 − 4(1) + 4( − 1) − 4
= −8
The resulting equation is 5x' 2 + 6x'y' + 5y' 2 − 8 = 0
Angle of rotation: tan2> = Ex3: (a) Determine the angle > through which the axes must be rotated to remove the xy-term in
(ii)

the equation xy = 2.
2B
A C
(1)
tan2@ = =∞
(b) Show that equation (1) represents a hyperbola that you will determine its geometric
cos@ −sin@ x''
6 formulas of rotation are:
2@ = 90° 4 5= 4 5
5 5 x' elements.

@ = 45° Sln: (a) A = 0, 2B = 1, C = 0, 2D = 0, 2E = 0, F = −2

(c) Sketch the graph of equation (1) by showing all the two sets of coordinate axes.
−
y' sin@ cos@ y''
B = 0.5 D=0 E=0
4 5 = A√2 √2
B 4 5 = A x'' y'' B
√2
1 1 x'' y''

Center of the conic: f(x, y) = xy − 2 = 0

x' x''

√2 √2 √2 f 'x = y = 0
y' 1 1 y'' (i)
3
f 'y = x = 0
x' = y' =
√2 √2 ∴ The center is at O(0, 0); h = 0 k=0
x'' y'' x'' y''
;

Angle of rotation: tan2> =

2B

5x’2 + 6x’y’ + 5y’2 − 8 = 0 tan2> = 0 0 = ∞

Substituting these values of x' and y' in (ii): A C

cos> −sin> x'

1 formulas of rotation are:
2> = 90° y= 4 5
5  +6  + 5  −8= 0
x

> = 45°
x'' y'' 2 x'' y'' 2
√2 √2 √2 √2
x'' y'' x'' y''

−
sin> cos> y'

54 5 +64 5 + 54 5 =8  y  = A√2 B 4 5 = A x'' y'' B

√2 √2
2 2 1 1 x'' y''
2 2 2 2 2
x'' 2x''y'' y'' x'' y'' x'' 2x''y'' y''2 x x'

=8
2 2 2
√2 √2
y'
√2
1 1
5x''2 10x''y'' 5y''2 6x''2 6y''2 5x''2 10x''y'' 5y''2

=8 x= y=
2
16x''2 4y''2
√2 √2
x' y' x' y'
2
;

Prepared by Jean Paul RUBAYIZA Page 263 Prepared by Jean Paul RUBAYIZA Page 264
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 xy  2
Substituting these values of x and y in (1):
6 A curve H has the equation x2  4y2  4x  8y  4  0.
  2
x’ y’ x’ y’
√2 √2
(a) Show that the curve H has a center of symmetry and determine its nature.
2
x’ 2 y’ 2 (b) Determine the vertices, the foci, the eccentricity, equations of directrices, and

x’  y’ 2  4
2 the equations of asymptotes of the curve H.
2 (c) Sketch the curve H.
 4 1
x  2 y2  6y  5
x’ 2 y’ 2 7 What can be said about the curve defined by the equation
4
(Equilateral hyperbola)
2x2  2xy  y2  2y  3  0.
8 A conic C has the equation:
V1, 2  a, 0  (  2, 0)
Vertices in new coordinates:
(a) Find the coordinates of the center of the conic
Focal length :
c  a2  b2  √4  4  √8  2√2
(b) Hence, write down the equation of the conic with respect to the new axes

Foci: F1, 2  c, 0  (  2√2, 0)

parallel to the old axes and passing through the center of the conic.

 (  2.8, 0)
(c) By rotation of the axes of symmetry of the conic, reduce the equation of the conic
C in standard form. Hence write down the nature of the given conic.
Length of the latus rectum:
LR   4
2b2 24
5.5. Conics and polar coordinates
 0
a 2
x' 2 y' 2
Asymptotes:
y' 2  x' 2
4 4 (1) Polar coordinates
y'  x' The Cartesian coordinates of a point P(x, y) can be
described in terms its distance from a fixed point

1 By translating the axes, using x  x'  h, y  y'  k, reduce each of the following
Exercise 5.10 and its direction (angle) from some fixed line
through this point.
• The fixed point O is called the pole and
(a) y2  6y  4x  5  0 (d) 2x2  3y2  4x  12y  20  0.
equations to its simplest standard form and state the nature of the locus.
• the directed line OX is called the polar axis
(b) x2  y2  2x  4y  20  0. (e) x2  2y2  4x  6y  8  0. r  x2  y2
(c) 3x2  4y2  12x  8y  4  0.
cosθ  r ⟺ x  rcosθ
x
2 Remove the first degree terms in each of the following equations by the method of

(a) x2  2y2  4x  6y  8  0 (c) 2x2  5y2  12x  10y  17  0.

completing the square.
sinθ  r ⟺ y  rsinθ
(b) 3x2  4y2  6x  8y  10  0. (d) 3x2  3y2  12x  12y  1  0.
y

(a) x2  4xy  2y2  4x  1  0 (c) (x  y)(2x  y  3)  0. tanθ  x ⟺ θ  arctan x

3 Find the coordinates of the center of each of the following conics. y y

(b) 2xy  3y2  4x  2y  0.

4 By translation and rotation of axes, reduce each of the following equation to its simplest ∴ x, y  (r, θ)

(a) 5x2  8xy  5y2  18x  18y  9  0.

standard form and plot its graph after stating its nature.

(b) 7x2  6√3xy  13y2  16.

(c) 25x2  14xy  25y2  64x  64y  224  0.

4x2  4xy  y2  8√5x  16√5y  0.

5 Simplify the following equation and state its nature.

Prepared by Jean Paul RUBAYIZA Page 265 Prepared by Jean Paul RUBAYIZA Page 266
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 Ex4: A curve C has a polar equation ρ2 = 4cos2θ. Transform its equation in
(a) (  2, 2√3) (b) (√3,  1)
Ex1: Convert the following Cartesian coordinates in polar coordinates

Sln: ρ2 = 4cos2θ
(c) (3, 4) Cartesian form.

Sln: (a) (  2, 2√3) , r  (  2)2  (2√3)  √4  12  √16  8 ρ2 = 4*cos2θ − sin2 θ+ (2)

(1) Substitute these expressions in (2):
x2 + y2 = 4 G4 2 25 − 4 2 2 5 H
2
2 2

cosθ   2 ' ρ = x2 + y2
x y

F
2 1

C ⇒ θ + = I − J
x y x y

cosθ = ρ = 2 2
sinθ  
2π
√3
4 x2 y2
x x

&
2 2
2√3 3 x y 4

F sinθ = ρ = 2 2 x + y = 4  x2 y2 
x2 y2 x2 y2

∴ ( − 2, 2√3) = (4,
x y

%
4 2 y y x2 y2
2π 2 2

∴ The required equation is:

3
) x y

(x2 + y2 2 = 4( x2 − y2 
(b) (√3, − 1) , r = (√3) + ( − 1)2 = √3 + 1 = √4 = 2
2

cosθ =
√3

C ⇒ θ=−
Ex5: Distance formula

sinθ = −
π
2 (a) Show that the distance between the points A(r1 , θ1 ) and B(r2 , θ2 ) is given by
d(A, B = r21 + r22 − 2r1 r2 cos(θ1 − θ2 )
1 6

∴ (√3, − 1) = (2, − )
2
π
6 π π

Sln: (a) A = (r1 , θ1 ) = (x1 , y1 ) = (r1 cosθ1 , r1 sinθ1 )

(b) Hence find the distance between the points P1 (3, 6 ) and P2 (6, 2 )

(c) (3, 4) , r = 32 + 42 = √25 = 5

B = (r2 , θ2 ) = (x1 , y1 ) = (r2 cosθ2 , r2 sinθ2 )
cosθ =
C ⇒ tanθ = ⟺ θ = arctan
3

sinθ =
4 4

∴ d(A, B = (x2 − x1 2 + (y2 − y1 2

5
4 3 3

∴ (3, 4) = (5, arctan )

5

= (r2 cosθ2 − r1 cosθ1 2 + (r2 sinθ2 − r1 sinθ1 2

4
3

Ex2: Convert the following polar coordinates in Cartesian coordinates = r22 cos2 θ2 + r21 cos2 θ1 − 2r1 r2 cosθ1 cosθ2 + r22 sin2 θ2 + r21 sin2 θ1 − 2r1 r2sinθ1 sinθ2

= r21 (cos2 θ1 + sin2 θ1  + r22 (cos2 θ2 + sin2 θ1  − 2r1 r2 (cosθ1 cosθ2 + sinθ1 sinθ2 
π 5π

Sln: (a) (2, 3π) = (2cos3π, 2sin3π) = ( − 2, 0)

(a) (2, 3π) (b) (5, ) (c) (√2, )
2 4

(b) (5, ) = (5cos , 5sin ) = (0, 5)

= r21 + r22 − 2r1 r2 cos(θ1 − θ2 ) which is the desired result.
π π π

) = (√2cos , √2sin ) = ( − 1, − 1) =
2 2 2
5π 5π 5π
(c) (√2,
4 4 4
π π
(b) P1 (3, ) ; P2(6, )
x2 + 3y2 = 3 d(P1 , P2  = 32 + 62 − 2(3)(6)cos  −  = 45 − 36cos  − 
Ex3: The Cartesian equation of a curve C is 6 2

Express its polar equation in the form ρ2 = b where a, b, c ∈ ℤ.

π π π
a 6 2 3

= 45 − 36   = √27
ccos2θ

C ≡ x2 + 3y2 = 3 (1)
1

ρ2  =3
Sln:

x = ρcosθ = 3√3
1 cos2θ 3 3cos2θ 2

 ρ2  2  =3
y = ρsinθ
2
4 2cos2θ
Remember that
cos2 θ = 2 (2
− cos2θ = 3
1 cos2θ

(ρcosθ)2 + 3(ρsinθ)2 = 3 ∴ ρ2 =
Subustitute these expressions in (1): ρ
2

ρ2 cos2 θ + 3ρ2 sin2 θ = 3

3

sin θ = 2
is the polar equation
2 cos2θ

ρ2 (cos2 θ + 3sin2 θ) = 3
2 1 cos2θ
of the curve C.

ρ2  +3 =3
1 cos2θ 1 cos2θ
2 2

Prepared by Jean Paul RUBAYIZA Page 267 Prepared by Jean Paul RUBAYIZA Page 268
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 Ex1: Determine the nature of the conic defined by the equation r = 4
Exercise 5.11 12
3cosθ

(b) (1, √3) (c) (  5, 12)

1. Convert the following cartesian coordinates into polar coordinates
Sln: Divide each term by 4:
r=4
(a) (2, 2) 12
Convert the following polar coordinates in Cartesian coordinates
(a) (3,  ) (b) (4,  ) =
3cosθ
5π π 2π 3
(c) (4, ).
(d) (13,  arctan 5 )
6 4 3 3

∴ e = 4 < 1, so the conic is an ellipse.

12 11π 1 cosθ
4
(e) (4, 3 ) 3

(a) y  ax (b) 9x2  16y2  144 (c) 9x  4y  36

2. Convert the following Cartesian equations in polar forms:

(d) y  4x (e) x3  xy2  2ay3  0

2 2

r=
Ex2: The polar equation of a curve C is given by
2 9
3. Change the following equations to rectangular coordinates 4 5cosθ

(a) θ  4 (d) r  2 sinθ

π 6 (a) Find the eccentricity and the nature of the curve.

(b) r2  2r(cosθ  sinθ)  7  0 (e) ρ  2

3
(b) Find the distance from the focus (pole) to the nearest directrix.

(c) ρ 
3cosθ (c) Convert the polar equation of the curve C in rectangular coordinates.
4
(d) Find the equations of asymptotes and directrices of the curve C in rectangular
1 cosθ
4. Find to the nearest tenth the distance between each of the following pair the points: coordinates.
π 5π

Sln: (a) r =
(a) (5, 45°) and (8, 90°) (c) (6, 12 ) and (8, 12 ) (e) Sketch the curve C.

) and (2,  3 )
9
5π π
(b) (3, 4 5cosθ

=
6 9
4
5
(2) Conics in polar coordinates
∴ e = > 1, so the locus is a hyperbola.
1 cosθ
4
5

(b) ep = 4 ⟺ p = 4 ⟺ p = 5 = 1.8
4
Derive the locus of a point P(r, θ) which moves so that 9 5 9 9

 e (a constant) ∴ The focus is 1.8 units from the directrix

OP 4

(c) r =
MP D
MP  NO  OQ  p  rcosθ 9(x2 + 10x − 16y2 + 81 = 0
Answer: 9

OP  r 4r − 5rcosθ = 9
9(x2 + 10x + 25 − 25 − 16y2 + 81 = 0
4 5cosθ
P(r, θ)
4 x2 + y2 − 5x = 9
M
∴ =e ⟺ =e 9(x2 + 2x + 25 − 16y2 + 81 − 225 = 0
4 x2 + y2 = 5x + 9
OP r

r = ep + ercosθ 9(x + 5)2 − 16y2 = 144

MP p rcosθ

r − ercosθ = ep 4 x2 + y2  = (5x + 92 −9 =1

2

r(1 − ecosθ = ep
(x 5)2 y2

16(x2 + y2 = 25x2 + 90x + 81 h = −5 , k = 0; center is at C(−5, 0)

r = 1 ecosθ
r 16

16x2 + 16y2 − 25x2 − 90x − 81 = 0 a = 4 , b = 3 (semi-axes)

ep

−9x2 + 16y2 − 90x − 81 = 0

p

9x2 − 16y2 + 90x + 81 = 0

If the directrix DD’ was to the right of the pole O,

r=
the equation would have the form
ep θ
1 ecosθ
If the directrix DD’ is parallel to the polar axis, the equation N O Q X

r=
takes the form D'
ep
1±esinθ

• If e < 1, the conic is an ellipse

The nature of the conic depends on the values of the eccentricity.

• If e > 1, the conic is a hyperbola

• If e = 1, the conic is a parabola

Prepared by Jean Paul RUBAYIZA Page 269 Prepared by Jean Paul RUBAYIZA Page 270
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 Ex3: Plot the graph of ρ 
Directrices : d1, 2 ≡ x  h  
2

Sln: ρ 
(d) Asymptotes a 1 sinθ

 0    3.2 x2  4y  1
e 2
(x 5)2 y2

∴ e  1, so the locus is a parabola.

a 4 16

h  0 , k  1;
1 sinθ

y  16 (x  5)2
16 9

∴ d1 ≡ x  5  3.2 ; d2 ≡ x  5  3.2
5
e 5

ρ  1 sinθ
9

4a  4
2 4

y   4 (x  5) d1 ≡ x  1.8 d2 ≡ x  8.2
2

ρ  ρsinθ  2 a1
3

x2  y2  y  2 The vertex is at V(0, 1)

(e) Vertices: V1, 2  (h  a, k)(  5  4, 0)
x2  y2  y  2
V1  (  1, 0) ; V2  (  9, 0) Focus: F  (h, k  a),
Foci: F1, 2  (h  c, k)  x2  y2   y  22 F  (0,  1  1)  (0, 1)
2

c  a2  b2  √16  9  √25  5 x2  y2  y2  4y  4
∴ F1, 2  (  5  5, 0) x2  4y  4 LR  4a  4
F1  (0, 0) ; F2 (  10, 0)
LR    2  4.5
2b2 29 9
a 4

Prepared by Jean Paul RUBAYIZA Page 271 Prepared by Jean Paul RUBAYIZA Page 272
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com
 (a) Find the locus of a point M which moves so that MF  MF'  10.
Exercise 5.12 5. On the rectangular coordinate system, your are given two points F(2, 2) and F'(8, 2).

1. Identify and sketch the locus of each of the following conic

(a) r  2 (d) r  3
4 10
(b) Sketch the curve
6. For each of the following conics given in parametric form, find its Cartesian equation
(b) ρ  2 (e) ρ  2
3cosθ 2cosθ

x  cosθ x  cosθ x  4tanθ

6 2 and plot its graph

(c) ρ  (a)  (b)  (c) 

sinθ 3sinθ

y  sin2 θ y  4sinθ y  3secθ

2

(d) x  1  4cosθ , y  2  4sinθ (e) x  t  t , y  t  t

1 cosθ
2. A curve C has a polar equation given by 1 1

ρ  5 4cosθ
4  1 represents:
9
x2 y2
(a) Find the eccentricity and identify the nature of the curve C. 7. (a) Show that the equation 29 c c
(b) What is the distance from the pole to the nearest directrix? (i) an ellipse if c is any constant less than 4.
(c) Convert the polar equation of the curve C in Cartesian (ii) a hyperbola if c is any constant between 4 and 29.
coordinates.
(c) If c  13, find the coordinates of the vertices, foci, equations of asymptotes,
(b) Show that the foci of the ellipse in part (a) are independent of the value of c.
(d) Find the equations of directrices to the curve C.
(e) Locate the foci of the curve C.
length of the latus rectum of the hyperbola and plot its graph.
3. Identify and sketch the curve r 
(f) Plot the graph of the curve C.
9 8. Find the Cartesian equation of the tangent line to each of the following curves at the

x  5cosθ x  3secθ
4 5sinθ specified point.
(a)  , θ (b)  , θ
y  3sinθ y  3tanθ
π π
Miscellaneous Exercise 5.13
3 4
9. The perimeter of an ellipse can be approximated by the formula:
4x2  y2  4  0
1. On the Cartesian plane, consider the curve C with equation:

P  2π
a2 b2
(a) Find the lengths of its semi-axes, vertices, foci, the eccentricity and the equations 2

of directrices of the curve. where a and b represent the length of the semi- major axis and semi-minor axis
(b) Sketch the curve C. respectively. Estimate the perimeter of the orbit of the planet Mercury, defined by
the equation:
5x2  y2  10x  4y  4  0
2. (a) Determine the standard form of the following flat shape:
 1245  1
x2 y2
1296
(b) Which shape is it?
 x  2  y  3   12
10. Given a conic with equation:
(c) Find the coordinates of the most points, the focal points, and the equation

3. On the Cartesian plane, consider the line L with equation x  1 and the point F with
of the focal line. Required:
(a) Determine the nature of the conic.
coordinates (3, 0) (b) Find the coordinates of the center of the conic.
(a) (i) Determine the Cartesian equation of the set C of all points M of the plane (c) Write down the equations of its axes of symmetry.
verifying the relation between the following distances:
dM, F  √3d(M, D) the line ∆ with equation 3y  2x  11.
(d) Calculate the coordinates of the points of intersection (A and B) of the conic and

(ii) Identify the nature of the set C and write down its characteristic elements. KKKKKKKL is always
perpendicular to KKKKKKKL
(e) Find the locus of a point M(x, y) which moves so that MA
(iii) Sketch the set C on the Cartesian plane.
MB.
y  rx, where r is a real parameter.
(b) Determine the number of points of the intersection of the set C and the line

4. Find the equations of tangents to the hyperbola x2  4y2  4  0 passing through

the point A(1, 1).

Prepared by Jean Paul RUBAYIZA Page 273 Prepared by Jean Paul RUBAYIZA Page 274
E-mail: jeyipi90@gmail.com E-mail: jeyipi90@gmail.com