CONIC SECTIONS

OBJECTIVES:
At the end of the lesson, the student is expected to be able to:
• define conic section
• identify the different conic section

Conic Section or a Conic is a path of point that moves so that its distance from a
fixed point (focus) is in constant ratio to its distance from a fixed line (directrix). The
constant ratio is called eccentricity.
Focus is the fixed point
Directrix is the fixed line
𝑐
Eccentricity is the constant ratio usually represented by (e) 𝑒 = 𝑎

The conic section falls into three (3) classes, which varies in form and in certain
properties. These classes are distinguished by the value of the eccentricity (e).
If e = 1, a conic section which is a parabola
If e < 1, a conic section which is an ellipse
If e > 1, a conic section which is a hyperbola
CIRCLES
OBJECTIVES:
At the end of the lesson, the student is expected to be able to:
• define circle

Circles are defined as a set of points that are equidistant (the same distance) from
a certain point; this distance is called the radius of a circle.
Here is the equation for a circle, where r is the radius:
Center (0,0): x2+y2=r2
Center (h,k): (x−h)2+(y−k)2=r2.
To solve for y in terms of x (for example, to put in the graphing calculator), we’d get:
Center (0,0): y= ±√𝒓𝟐 − 𝒙𝟐
Center (h,k): y= ±√𝒓𝟐 − (𝒙 − 𝒉)𝟐 + 𝒌

Sometimes we have to complete the square to get the equation for a circle. We
learned how to complete the square with quadratics under factoring and completing the
square (Algebra lesson). Note that, after completing the square, we may not necessarily
get a circle if the coefficients of x2 and y2 are not both positive 1.

General Equation. The general equation of second degree is of the form

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

If A = C and B = 0, then the equation becomes

Ax2 + Ay2 + Dx + Ey + F = 0

Dividing through by A and then completing the squares in x and y. If after completing
the squares in x and y, the right-hand member if the equation is zero, the locus is a
point; if negative, there is no locus.

𝑫 𝑬 𝑭
Ax2 + Ay2 + 𝑨x + 𝑨y + 𝑨 = 0

x2 + y2 + dx + ey + f= 0.

Example 1: Reduce to standard form of the circle whose equation is 4x2 + 4y2 -
20x + 4y - 55 = 0

Solution:

Divide through by 4 [ 4x2 + 4y2 - 20x + 4y - 55 = 0]

55
x2 + y2 - 5x + y - =0
4

Transpose the constant and complete squares in x and y

25 1 55 25 1
x2 - 5x + + y2 +y + 4 = + +4
4 4 4
5 1 9
(x - 2)2 + (y + 2)2 = (2)2
5 1 9
The center is at c (2 , − 2) and the radius r is 2.

Example 2.
Find the center and radius of the following circle: x2+y2+8y−9=0
Solution:
x2+y2+8y = 9
x2+(y2+8y+ ?_) = 9 + __?_
x2+(y2+8y+ (4)2) = 9 + (4)2
x2+(y+4)2 = 52
The equation is now in circle form. Note that x2 is the same as (x−0)2, so this
equation is a circle with center (0, −4) and radius 5.
Example 3:
Find the center and radius of the following circle: x2+y2−6x−12y−55=0
Solution:
x2+y2−6x−12y = 55
(x2− 6x+ ___) + (y2 - 12y+___) = 55 + ____
(x2− 6x+ _(3)2_) + (y2 - 12y+_(6)2_) = 55 + __(3)2_+_(6)2__
(x - 3)2 + (y - 6)2 = 55+ 9+36
(x - 3)2 + (y - 6)2 = 100
(x - 3)2 + (y - 6)2 = 102
The equation is now in circle form! This equation is a circle with center (3,6) and
radius 10.

Example 4:
Write the equation of the line that is tangent to the circle (x−3)2+(y+2)2=61 at the point
(−2, −8).

Solution:
A line tangent to a circle means that it touches the circle at one point on the
outside of the circle, at a radius that is perpendicular to that line:

For this problem, since we only have one point on the tangent line (−2, −8), we’ll have to
get the slope of the line to get its equation. Remember your lesson in Coordinate
System and Graphing Lines that perpendicular lines have slopes that are opposite
reciprocals of each other.
 We can get the slope of the line that
connects the center of the circle (3, −2) and
the point on the tangent line (−2, −8), and
then take the negative or opposite
reciprocal to get the slope of the tangent
line.
The slope of the line that contains (3,−2)
𝑦2−𝑦1 −8−(−2) −6 6
and (−2,−8) is = = − −5 =
𝑥2−𝑥1 2−3 5

Since the tangent line is perpendicular to

5
this line, its slope is .
6

Using this slope and point (–2, –8), we can

use either the slope-intercept or point-slope
method to get the equation; let’s use the
slope-intercept:
5 5
𝑦=− 𝑥+𝑏 −8 = − (−2) + 𝑏
6 6
10 29
𝑏 = −8− =−
6 3

The equation of the tangent line is y =

5 29
−6 𝑥 − 3

Example 5: Find the equations of the circle passing through

P1 (-3,6), P2(-5,2), P3 (3, -6).
Solution:
Substitute the coordinates of P1, P2, P3 in x2 + y2 + dx + ey + f= 0, this will yield to
-3d + 6e + f = -45 eq.1
-5d + 2e + f = -29 eq.2
3d – 6e + f = -45 eq.3
Solving the three equations simultaneously, we get
d = -4; e = -2; f = -45
Therefore, the equation of the circle is
x2 + y2 - 4x – 2y - 45= 0.

Exercise:
1. Find the equations of the circle with center (1,4) and the radius is 5.
Ans. x2+y2 – 2x – 8y – 8 = 0
2. Find the equations of the circle with center (2, -5) and passing through (3,4).
Ans. x2+y2 + 4x + 10y - 53 = 0
3. Find the equation of a circle through the intersection of 4x +y -4 = 0 and x – y – 6
= 0 with center at (-1, -3)
4. A circle has its center on the line 3x – 2y – 22 = 0 and tangent to the y-axis at
(0,1). Find its equation
Ans. x2+y2 – 16x – 2y + 1= 0
5. Find the points of intersection of the line x – 2y + 2 = 0 and the circle x2+y2 +6x +
8 = 0.
Ans. (-2,0) and (-6, -2)

PARABOLA
OBJECTIVES:
At the end of the lesson, the student is expected to be able to:
• describe parabola
• convert general form to standard form of equation of parabola and vice versa.
• give the different properties of a parabola and sketch its graph

THE PARABOLA (e = 1)
A parabola is the set of all points in a plane, which are equidistant from a fixed
point and a fixed line of the plane. The fixed point called the focus (F) and the fixed line
the directrix (D). The point midway between the focus and the directrix is called the
vertex (V). The chord drawn through the focus and perpendicular to the axis of the
parabola is called the latus rectum (LR).
PARABOLA WITH VERTEX AT THE ORIGIN, V (0, 0)
Let: D - Directrix
F - Focus
2a - Distance from F to D
LR - Latus Rectum = 4a
(a, 0) - Coordinates of F
Choose any point along the
parabola
So that,
𝑃𝐹 𝑃𝐹
=𝑒=1 =1
𝑃𝑀 𝑃𝑀

√(𝑥 − 𝑎)2 + (𝑦 − 0)2 = [𝑥 − (−𝑎)]

√𝑥 2 − 2𝑎𝑥 + 𝑎2 + 𝑦 2 = 𝑥 + 𝑎
Squaring both sides:
X2-2ax+a2+y2 = (x+a)2
X2-2ax+a2+y2 = x2+2ax+a2
-2ax+y2 = +2ax
y2 = +4ax

General Form of the Parabola

Consider an equation of the second degree in which the x 2 and xy terms are
missing. Such equation would be of the form.

Cy2 + Dx + Ey + F = 0

Dividing through by C

𝑫 𝑬 𝑭
y2 + 𝑪 x + 𝑪y + 𝑪 = 0

𝐷 𝐸 𝐹
Let d = 𝐶 x, e = 𝐶 y and f = 𝐶

The equation may thus be written: y2 + dx + ey + f = 0,

Transpose the x and y terms to the right complete square and simplify

y2 - ey = - dx – f
𝑒2 𝑒2
y2 +ey + = -dx – f + 4
4
𝑒 1 𝑒2
(y +- 4)2 = - d [x +2(f - )]
4

Therefore, Cy2 + Dx + Ey + F = 0.

Equations of parabola with vertex at the origin V (0, 0)

𝑦 2 = +4𝑎𝑥 𝑦 2 = - 4ax

𝑥 2 = +4𝑎𝑦 𝑦 2 = - 4ax
PARABOLA WITH VERTEX AT V (h, k)
We consider a parabola whose axis
is parallel to, but not on, a coordinate
axis. In the figure, the vertex is at (h,
k) and the focus at (h+a, k). We
introduce another pair of axes by a
translation to the point (h, k). Since
the distance from the vertex to the
focus is a, we have at once the
equation
y’2 = 4ax’
Therefore, the equation of a
parabola with vertex at (h, k) and
focus at (h+a, k) is
(y – k)2 = 4a (x – h)

Equations of parabola with vertex at V (h, k)

(𝑦 − 𝑘)2 = +4𝑎(𝑥 − ℎ) (𝑦 − 𝑘)2 = −4𝑎(𝑥 − ℎ)

 (𝑥 − ℎ)2 = +4𝑎(𝑦 − 𝑘) (𝑥 − ℎ)2 = −4𝑎(𝑦 − 𝑘)

Standard Form General Form

(y – k)2 = 4a (x – h) y2 + Dy + Ex + F = 0
(y – k)2 = - 4a (x – h)
(x – h)2 = 4a (y – k) x2 + Dx + Ey + F = 0
(x – h)2 = - 4a (y – k)

Example 1.

1. Find the equation and draw the general appearance of the parabola with vertex at (-
1,4) and y = 5 as directrix.
Solution:
Comparing the equation of the directrix with the ordinate of the vertex, it is evident that
the directrix is above the vertex, therefore, the parabola opens downward and is the
type of
(x – h)2= -4a (y-k)
The distance a from the vertex to the directrix is equal to (5, -4) or 1, the length 4a of the
latus rectum is 4 and h and k are -1 and 4 respectively.

Substituting (x – h)2= -4p (y - k), we obtain the equation of the parabola to be

(x – (-1))2= -4a (y - 4),
 X2 + 2x +4y – 15 = 0
General appearance of the parabola

Exercises
1. Find the equations of a parabola with axis parallel to the x-axis and passing
through (3/2, 1), (5,0) and (-1, 2). (Ans. y2-2x-8y+10 = 0)
2. Find the equation of a parabola with vertex on the line y = 3x, axis parallel to the
y-axis and passing through (-7,13) and (5,1). (Ans. x2-2x-4y -11 = 0)
3. Water spouts from a horizontal pipe 40 feet above the ground and 3 feet below
the line of the pipe, the water trajectory is at a horizontal distance of 16 feet from
the vertical line through the end of pipe. How far from the vertical line will the
stream of the water hit the ground? (Ans. 16√2 feet)
4. A parkway 80 feet wide is spanned by a parabolic arc 100 feet long along the
horizontal. If the parkway is centered, how high must the vertex of the arch be in
5
order to give a minimum clearance of 5 meters over the parkway. (Ans. 559 feet)
ELLIPSE
OBJECTIVES:
At the end of the lesson, the student is expected to be able to:
• define ellipse
• give the different properties of an ellipse with center at (0,0)
• identify the coordinates of the different properties of an ellipse with center at
(0, 0)
• sketch the graph of an ellipse

THE ELLIPSE (e < 1)

An ellipse is the set of all points P in a plane such that the sum of the distances
of P from two fixed points F’ and F of the plane is constant. The constant sum is equal
to the length of the major axis (2a). Each of the fixed points is called a focus (plural
foci).
The following terms are important in drawing the graph of an ellipse:
Eccentricity measure the degree of flatness of an ellipse. The eccentricity of an ellipse
should be less than 1.
Focal chord is any chord of the ellipse passing through the focus.
Major axis is the segment cut by the ellipse on the line containing the foci a segment
joining the vertices of an ellipse
Vertices are the endpoints of the major axis and denoted by 2a.
Latus rectum or latera recta in plural form is the segment cut by the ellipse passing
through the foci and perpendicular to the major axis. Each of the latus rectum can
2𝑏2
be determined by: LR = 𝑎

Properties of an Ellipse:
1. The curve of an ellipse intersects the major-axis at two points called the vertices.
It is usually denoted by V and V’.
2. The length of the segment VV’ is equal to 2a where a is the length of the semi-
major axis.
 3. The length of the segment BB’ is equal to 2b where b is the length of the semi-
minor axis.
4. The length of the segment FF’ is equal to 2c where c is the distance from the
center to the foci.
5. The midpoint of the segment VV’ is called the center of an ellipse denoted by C.
6. The line segments through F1 and F2 perpendicular to the major – axis are the
latera recta and each has a length of 2b2/a.
7. The relationship of a, b and c is given by; a2 = b2 + c2 where,
a > b.
8. c = a e

ELLIPSE WITH CENTER AT ORIGIN C (0, 0)

ELLIPSE WITH CENTER AT ORIGIN C (0, 0)

d1 +d2 = 2a
Considering the triangle F’PF
d3 +d4 = 2a
d3 = 2a - d4
√[𝑥 − (−𝑐 )]2 + 𝑦 2 = 2a - √[𝑥 − 𝑐)2 + (𝑦 − 0)2
 (√[𝑥 − (−𝑐 )]2 + 𝑦 2 = 2a - √[𝑥 − 𝑐)2 + (𝑦)2 )2

X2+2cx+c2+y2 = 4a2 – 4a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2 +𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2

2cx = 4a2 -4a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2 – 2cx

2cx+2cx-4a2 = -4a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2

4cx-4a2 = -4a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2

a2-cx = a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2

(a2-cx = a√𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2 )2

a4 – 2a2cx + c2x2 = a2 (x2 - 2cx + c2 + y2)
a4 – 2a2cx + c2x2 = a2 x2 - 2a2cx + a2c2 + a2y2
a2(a2-c2) = x2(a2-c2) + a2y2
But: c2 = a2 – b2
b2 = a2 – c2
a2 b2 = x2b2+a2y2
1
[a 2 b 2 = 𝑥 2 𝑏2 + 𝑎2 𝑦 2 ]
𝑎2 𝑏 2
𝑥2 𝑦2
+ = 1 standard form.
𝑎2 𝑏2

Equations of ellipse with center at the origin C (0, 0)

𝑥2 𝑦2 𝑥2 𝑦2
+ =1 + =1
𝑎2 𝑏 2 𝑏 2 𝑎2
ELLIPSE WITH CENTER AT C (h, k)
 If the axes of an ellipse are parallel to the coordinate axes and the center is at (h,k),
we can obtain its equation by applying translation formulas. We draw a new pair of
coordinate axes along the axes of the ellipse. The equation of the ellipse referred to the
new axes is
𝑥′2 𝑦′2
+ 2 =1
𝑎2 𝑏
The substitutions x’ = x – h and y’ = y – k yield
ELLIPSE WITH CENTER AT (h, k)

(𝑥 − ℎ)2 (𝑦 − 𝑘)2 (𝑦 − 𝑘)2 (𝑥 − ℎ)2

+ =1 + =1
𝑎2 𝑏2 𝑎2 𝑏2

Exercises:
1. Write the equation of the ellipse with center at (2,1) with major axis = 10 and parallel
to the x-axis, and with minor axis = 8. (Ans. 16x2 + 25y2 – 4x – 50y – 311 = 0)
2. Sketch the ellipse 9x2 + 25y2 = 225
3. Find the coordinates of the foci, the end of the major and minor axes, and the ends of
each latus rectum. Sketch the curve.
𝑦2 𝑥2 (𝑥−2)2 (𝑦−3)2
a. + =1 b. + =1
25 16 16 9

4. Reduce the equations to standard form. Find the coordinates of the center, the foci,
and the ends of the minor and major axes. Sketch the graph.
a. x2 + 4y2 – 6x –16y – 32 = 0
b. 16x2 + 25y2 – 160x – 200y + 400 = 0
c. 3x2 +2y2 – 24x + 12y + 60 = 0
d. 4x2 + 8y2 + 4x + 24y – 13 = 0
5. The arch of an underpass is a semi-ellipse 6m wide and 2m high. Find the clearance
at the edge of a lane if the edge is 2m from the middle. (Ans.1.49m)
y

●( 2

C x
2
6
5. The earth’s orbit is an ellipse with the sun at one focus. The length of the major
axis is 186,000,000 miles and the eccentricity is 0.0167. Find the distances from
the ends of the major axis to the sun. These are the greatest and least distances
from the earth to the sun.
(Ans. least dist. = 91,446,900 miles greatest dist. = 94,553, 100 miles)

Least distance
Least

C F
s●

Greate
Greatest distance

7. A hall that is 10 feet wide has a ceiling that is a semi-ellipse. The ceiling is 10 feet
high at the sides and 12 feet high in the center. Find its equation with the x-axis
horizontal and the origin at the center of the ellipse. (Ans. 4x2+25y2-100 = 0)
HYPERBOLA
OBJECTIVES:
At the end of the lesson, the student is expected to be able to:
• give the properties of hyperbola.
• write the standard and general equation of a hyperbola.
• sketch the graph of hyperbola accurately.

THE HYPERBOLA (e > 1)

A hyperbola is the set of points in a plane such that the difference of the
distances of each point of the set from two fixed points (foci) in the plane is constant.
The equations of hyperbolas resemble those of ellipses but the properties of
these two kinds of conics differ considerably in some aspects.
To derive the equation of a hyperbola, we take the origin midway between the
foci and a coordinate axis on the line through the foci.

The following terms are important in drawing the graph of a hyperbola

Transverse axis is a line segment joining the two vertices of the hyperbola.
Conjugate axis is the perpendicular bisector of the transverse axis.
General Equations of a Hyperbola
1. Horizontal Transverse Axis: Ax2 – Cy2 + Dx + Ey + F = 0
2. Vertical Transverse Axis: Cy2 – Ax2 + Dx + Ey + F = 0
HYPERBOLA WITH CENTER AT THE ORIGIN C (0,0)

We denote the foci F’(-c,0) and F (c,0) and the difference between a point of the
hyperbola and the foci by the positive constant 2a. Then from the above definitions
|F’P | - |FP| = 2a or |F’P| - |FP| = -2a
Depending on whether the point P(x,y) of the hyperbola is to the right or left of
the y-axis. We combine these equations by writing
|F’P | - |FP| = 2a
√(𝑥 + 𝑐)2 + 𝑦 2 - √(𝑥 − 𝑐)2 + 𝑦 2 = ±2a
Now, transposing the second radical, squaring and simplifying we obtain,
cx – a = ±2a√(𝑥 − 𝑐)2 + 𝑦 2
Squaring the members of this equation and simplifying, we get
(C2 – a2)x2 –a2y2 = a2(c2 – a2)
Then letting b2 = c2 – a2 and dividing by a2b2, we have
 𝑥2 𝑦2 - if foci are on the x-axis
− =1
𝑎2 𝑏 2

𝑦2 𝑥2 - if foci are on the y-axis

− =1
𝑎2 𝑏 2 -
The generalized equations of hyperbolas with axes parallel to the coordinate
axes and center at (h, k) are

(𝑥 − ℎ)2 (𝑦 − 𝑘)2 - if foci are on an axis parallel to the x-

− =1
𝑎2 𝑏2 axis
(𝑦 − 𝑘)2 (𝑥 − ℎ)2 - if foci are on an axis parallel to the y –
− =1
𝑎2 𝑏2 axis

Equations of the asymptote

𝑥2 𝑦2 𝑏
− =1 y = ±𝑎𝑥
𝑎2 𝑏2
𝑦2 𝑥2 𝑎
− =1 y = ±𝑏𝑥
𝑎2 𝑏2
(𝑥−ℎ)2 (𝑦−𝑘)2 𝑏
− =1 y - k = ± 𝑎 (𝑥 − ℎ)
𝑎2 𝑏2
(𝑦−𝑘)2 (𝑥−ℎ)2 𝑎
− =1 y - k = ± 𝑏 (𝑥 − ℎ)
𝑎2 𝑏2

Important Relations
1. c > a, c > b, (a = b, a < b, a > b)
2. c2 = a2 + b2 or ae = √𝑎2 + 𝑏2
𝑐 √𝑎2 +𝑏 2
3. e = 𝑎 ; or e = 𝑎
𝑎 𝑎
4. Directrix = = =
𝑒 𝑎√𝑎2 +𝑏2
2𝑏2
5. LR = 𝑎
Example1.
Sketch the graph of the following hyperbola.

Solution:
The first step here is to simply compare our equation to the standard form of the
hyperbola and identify all the important information. For reference purposes here is the
standard form of the hyperbola that matches the one we have here.

Comparing our equation to this we can see we have the following information.
h = 2, k = 0, a = 3, and b = 4
Because the y term is the positive term, we know that this hyperbola will open up and
down. With the information we found in the first step we can see that the center of the
hyperbola is (2,0). The vertices of hyperbola are: (2, −4) and (2,4).

The equations of the two asymptotes are,

Here is a sketch of the hyperbola including the points and asymptotes we found above.
Example 2
Write the equation into the standard form of the equation of the hyperbola.

The first step is to make sure the coefficient of the x2 and y2 is a one. The x2 has a
coefficient of 4 and the y2 has a coefficient of -1. What we will do is factor a 4 out of
every term involving an x and a -1 out of ever term involving a y. Doing that gives,

Now let’s get started on completing the square. First, we need one-half the coefficient of
the x and y term, square each and the add/subtract those numbers in the appropriate
places as follows,

Next, we need to factor the x and y terms and add up all the constants.

To finish things off we’ll first move the 36 to the other side of the equation.

To get this into standard form we need a one on the right side of the equation. To get
this all we need to do is divide everything by 36 and we’ll do a little simplification work
on the x term.
Exercises
1. Find the equation of the hyperbola with foci (5, 2) and (-1, 2) whose transverse
(𝑥−2)2 (𝑦−2)2
axis is 4 units long. (Ans. − = 1)
4 5
2. Find the coordinates of the center, foci, and vertices, and the equations of the

asymptotes of the graph . Then graph the equation.

3. Find the coordinates of the center, foci, and vertices, and the equations of the
asymptotes of the graph of 25x2 – 16y2 + 250x + 32y + 109 = 0. Then graph the
equation.
4. A pair of buildings on a college campus are shaped and positioned like a portion
of the branches of the hyperbola 225x2 – 400y2 = 90,000 where x and y are in
meters. How far apart are the buildings at their closest point? (Ans. The vertices
are at (-20, 0) and (20, 0). So, at their closest point, the buildings are 40 meters
apart.)
5. Find the equation of the hyperbola which satisfies the given conditions
a. Center (0,0), transverse axis on the x-axis, a = 4, latus rectum 32. (Ans. 4x2-
y2 = 64)
b. center at the origin, transverse axis on the, distance between foci 2√97 latus
rectum 32/9. (Ans. 16x2 -81y2 = 1,296)
c. Center (0,0), transverse axis along the y-axis, eccentricity √53 / 2, distance
between foci 2 √53 . (Ans. 49y2 – 4x2 = 196)
d. Center (0,0), transverse axis along y-axis, slope of asymptote numerically
equal to nine-tenths of the conjugate axis, one vertex at (0, -9). (Ans. 5y2 –
81x2 = 405)
e. Center (0,0), transverse axis on the x-axis, conjugate axis 24, (2/3 √145 , 1)
is appoint on the curve. (Ans. 9x2 – 4y2 = 576)