ANALYTIC GEOMETRY

Rectangular Coordinate or Cartesian System

1. Point is denoted by the ordered pair x and y as (x, y)

2. X – coordinate - called abscissa
The distance from the origin to the projection of the point unto the x-axis
3. Y - coordinate - called ordinate
The distance from the origin to the projection of the point unto the y-axis
4. Origin - The intersection of the x and y – axis

Distance between Two Points P1 (X1, Y1) and P2(X2, Y2)

d= √ ¿ ¿

Area of Polygon (Non-overlapping) of n-sides Given Vertices

Given vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) , … .. ( x n , y n ) oriented counterclockwise

1
A=
2
[ ( X 1 Y 2+ X 2 Y 3+ … X n Y 1 )−( Y 1 X 2 +Y 2 X 3 +… Y n X 1 ) ]
Sample Problem

Find the area of polygon joining the points (3, 0), (2, 3), (-1, 2), (-2, -1) and (0, 2).
Solution:
Note: The points are given already in the counterclockwise orientation
Thus:
A=
[
1 3 2 −1 −2 0 3
2 0 3 2 −1 −2 0 ]
A - 1/2 [(9 +4+ 1+4+ 0) - (0 - 3 - 4 - 0 - 6)]
A = 3 1/2 square units
 Division of Line Segment
Let P (x, y) be a point on the line joining P1 (Xl, Yl) and P2 (X2, Y2) and located in such a way that segment P1 P is a
given fraction k P1P2, that is P1 P = P1P2.

x = x1 + k (x2 - x1)
Y = Yl + k (Y2 -Yl)

If k = 1/2, then, formula above becomes a

X0 = 1/2 (x1+ x2); y0 = 1/2 (yl + y2)
Sample Problem:
Find the coordinate of the point which 1/3 of the way from A (l, -l) and B (7, 5).
Solution:
X = x1 + k(x2-x1) y = y1 + k(y2-y1)
X = 1+1/3 (7-1) y = -1+1/3(5+1)
X=3 y=1
Thus, point required is P (3, 1) Answer

Projection of a Line Segment on the Coordinate Axes

5. The directed segment A1A2 is the projection of the segment P1P2 on the x-axis
6. The directed segment B1B2 is the projection of the segment P1P2 on the y-axis
Obviously, A1A2 = x2 - x1
B1B2 = y2 - yl

Sample problem:
Given two points P1 (2, -1) and P2(6, 5), find the projections of P1P2 on the x and y axis
Solution:
a) Projection on x-axis = x2 - x1
= 6-2 =4
b) Projection on y-axis = y2 - yl
= 5-(-1) =6
Slope, Parallel and Perpendicular Lines

Line L

Definition
1. The inclination (θ ) of a line L is defined as the least positive angle made by the line with the positive x-axis.
2. The slope (m) of a line L is defined as the tangent of its inclination,
rise y 2− y 1
m=tan θ m= =
run x 2−x 1
Condition for Parallel
1. Two lines L1 and L2 are perpendicular if their slopes are equal (ml = m2)
2. Condition for Perpendicularity
Two lines L1 and L2 are perpendicular with each other if the slope of one equal to the negative reciprocal of the
slope of the other
−1
m 1= ∨m1 m2=−1
m2
Notes:
1. If the Line is inclined to the left, 90o< θ <180o and m < 0.

2. If the line L is inclined to the right, 0 < θ < 90o and m > 0

3. If the line parallel to the x-axis, θ = 0 and m = 0

 4. If the line is parallel to the y-axis, θ = 90o and m is undefined

Angle between Two Concurrent Lines

Let α and β be the inclinations of lines Ll and L2 respectively and let θ be the angle between the two lines.

θ=β -α ; m1 = tan
α
y
tan θ = tan ( β - α )
tan β−tan α
tan θ= m2=tan β
1+ tan β−tan α
m2−m1
tanθ=
1+ m1 m2

Sample Problem:
Given three points A (-1, 1), B (5, -1), C (4, 3), find the angle made by segments AB and AC.
Solution:
First, find the slope of AB and AC
−1−1 −1 3−1 2
m AB= = ; m AC = =
5+1 3 4+ 1 5

2 1
+
m2−m1 5 3
tanθ=¿ = ¿
1+ m1 m2
1−
2 1
5 3 ()
11
tanθ=
13
−1 11
θ=tan , θ=40.236 ° Answer
13

Locus - the curve traced by an arbitrary point as it moves in a plane is called locus of a point
- the locus of an equation is a curve containing only those points whose coordinates satisfy the equation.

EQUATION OF A STRAIGHT LINE

Line - is a locus of points which has constant slope.
Theorems:
5. Every straight line can be represented by a first-degree equation.
6. The locus of an equation of the first degree is always a straight line.
General Equation of a Line
Ax + By + C = O ; A, B, C are constants ; A and B, not zero at the same time

Standard Equation of a Line

1. Two Point Form

By similarity of triangles
 y 2− y 1
y− y 1= ( x−x 1 ¿
x2− x1

2. Point-slope Form y
y 2− y 1
ln (1) replacing by m,
x 2−x 1
y− y 1=m(x−x 1)

m=tan θ
0 ° ≤θ ≤ 180 °

3. Slope-Intercept Form
y=mx+b
where m = slope
b= y intercept

4. Two-Intercept Form
x y
+ =1
a b
Where: a = x intercept
b = y intercept

5. Normal Equation of a Straight Line

Given ρ = normal intercept
= segment from the origin
perpendicular to the required line
θ = normal angle
= inclination of the normal intercept

x cosθ +ysinθ = ρ
Reduction to Normal Form:
Given the line Ax + By + C = 0

The normal form is

A B C
x+ y+ =0
± √ A +B
2 2
± √ A +B
2 2
± √ A2 + B2

Note: The sign of the radicand must be chosen such that the last term will become negative since ρ>0

Special Cases of a Straight Line

A. Equation of the x-axis y=0
Equation of a horizontal line y=b where b is a constant
B. Equation of the y-axis x=0
 Equation of a vertical line x=a where a is constant

Sample Problems:

1. Find the equation of a line passing through (-1, -3) and (2, 4)
Solution:
y 2− y 1
Using two-point form formula: y− y 1= ¿
x2− x1
Where x 1=−1 x2 =2
y 1=−3 y 2=4
4+3
y +3= ( x +1)
2+1
Simplifying, 7 x−3 y−¿ 2=0 Answer

2. Find the equation of the line having a slope of 1/3 and passing through the point (5, 1).
Solution:
Using the point-slope form: y – y 1=m ( x−x 1 )
1
Where, x 1=5 , y 1=1 , m=
3
1
y−1= (x−5)
3
Simplifying, x- 3y -2 = 0 Answer

3. Find the equation of the line with slope of — 2 and y intercept of 3.

Solution:
Using slope-intercept form: y = mx + b, where, m = -2, b = 3
y = -2x + 3 or 2x + y - 3 = 0 Answer
4. Find the equation of a line with x and y intercept equal to 3 and 5 respectively
Solution:
x y
Using the two-intercept form: + =1
a b
where a = x intercept = 3
x y
b = y intercept = 5 + =1
3 5
Simplifying, 5x + 3y – 15=0 Answer

5. Find the equation of a line passing through (2, -1) and parallel to the line 12x - 4y + 5 = 0 Solution:
Since the required line is parallel to the given line their slopes are equal
12 5
For the given line, 12x – 4y + 5 = 0 or y= x+
4 4
Here slope m = 3
Using the same slope m = 3
y− y 1=m ( x−x 1 ) where x 1=2 y 1=−1 m=3
y +1=3 ( x−2 )
Simplifying, 3x – y – 7 = 0 Answer

6. Find the equation of a line having normal angle = 60 0 and normal length 1
Solution:
Normal form: x cos θ+ y sin θ= ρ
Where θ=60 ° , ρ=1
x cos θ+ y sin θ=1
Simplifying, x+
1 √ 3 y=1
2 2
x + √ 3 y −2=0 Answer

7. Reduce the equation 3x - 4y + 10 = 0 to normal form

Solution:
A=3 B = -4 C =10
3 x−4 y +10
Normal form is
± √ [3 + (−4 ) ]
2 2
 Choose the - sign of radical because B is negative (to make C negative).
−3 4
Thus x + y−2=0Answer
5 5

Distance Between Parallel Lines

Let the parallel lines be given by the equations:
L1 : Ax+ By +C 1=0
L1 : Ax+ By +C 2=0
The distance d between the two

Lines is given by the formula:

C 2−C 1
d=
± √ A +B
2 2

Sample Problems:

1. Find the distance from point (3, 1) to the line 3x - 4y - 3 = 0

Solution:
Here, A = 3, B = -4, C = -2 Po ( x o , y o ) ↔ ( 3 ,−1 )

Ax o+ By 0 +C 3 ( 3 ) + (−4 ) (−1 )−4

using the formula d= =
± √ A +B ± √3 + 4
2 2 2 2

d = 2 units (the point (3, - 1) and the origin are on the opposite side of the line)

2. Find the distance between parallel lines 8x + 15y + 18 = 0 and 8x + 15y + 1 = 0

Solution:
A = 8, B = 15, C2 = 18, C1 = 1
C 2−C 1 18−1
D= = =1 unit Answer
± √ A +B ± √ 8 +15
2 2 2 2

Distance from a Point to a Line

Ax o+ By 0 +C
d=
± √ A +B
2 2

The directed distance from a point P( x o , y o) to a line Ax +By + C = 0 is given by the formula:
where the sign of the radical is same as sign of B (or C when B = 0) so that,

Remarks:
1. if d > 0, the origin and P lie on opposite sides of the given line
2. if d < 0, the origin and P lie on the same side of the line

Notes:
Regardless of the location of the point Po ( x o , y o), the distance being always positive the formula can be expressed using
the absolute value as:
| Ax o +By 0 +C|
d=
√ A2 + B2
Line Through the Intersection of Two Lines
Let Ax + By + C = 0and
Dx + Ey + F = 0 be two intersecting lines, where A, B, C, D, E and F are constant and
 A = B = C = D = E = F≠ 0
The equation of the family of lines passing through the intersection of the two given lines is given by,
(Ax + By + C) +k (Dx + Ey + F) = 0
Where k is an arbitrary constant

Sample Problem:
1. Find the equation of the line passing through the intersection of the lines x - y + 5 = 0 and x +
5y – 4 = 0 and passing through the point (1, 0).
Solution:
The equation of the family is (x - y + 5) + k (x + 5y - 4) = 0
Since the required line is a member of the family, substitute the point (1, 0) in the locus of the family and then solve
for k
[(l) - 0 + 5] + k [l + 5(0) -4] = 0
6 – 3k = 0, k = 2
Substituting k in the equation of the family of lines above,
(x – y + 5) + 2 (x +5y – 4) = 0
Simplifying, 3x + 9y - 3= 0
x + 3y – 1 = 0 Answer

INTERCEPT OF A CURVE
x- intercept - directed distance from the origin to the point where the curve crosses the x-axis
To find the x intercept of a curve, set y = 0, then solve for x.
y - intercept - the directed distance from the origin to the point where the curve crosses the y- axis. To find the y intercept
of a curve, set x = 0, then solve for y symmetry
1. if the equation of a curve does not change upon replacement of y by - y, then the locus is symmetric with respect to
the x-axis
f (x, - y) = f (x,y) = 0
2. if an equation of a curve does not change upon replacement of x by - x, then the locus is symmetric with respect
to the y-axis
f (-x, y) = f (x, y) = 0
3. if an equation of a curve does not change upon replacement of x by - x and y by - y, then the locus is symmetric
with respect to the origin.
f (- x, - y) = f (x, y) = 0

ASYMPTOTE - a straight line which the curve f(x, y) = 0 approaches indefinitely near as its tracing point approaches
to infinity.
4. To find the vertical asymptote, solve the equation for y in terms of x and set the linear factors of the denominator
equal to zero.
5. To find the horizontal asymptote, solve the equation for x in terms of y and set the linear factors of the
denominator equal to zero.

CIRCLE
Circle is the locus of a point which moves so that it is always equidistant from a point
Note: fixed point is called the center & fixed distance is called the radius

Equation of a Circle
A. In normal form
Consider a circle of radius r with center at C (h, k)
Let P(x, y) be a point in the circle

By Pythagorean Theorem
¿ STANDARD FORM

Center at the origin C (0, 0)

2 2 2
x + y =r
 B. General form

Expanding the form (x – h)2 + (y – k)2 = r 2 becomes

x2 + y2 - 2xh - 2ky + h2 + k 2 - r 2 = 0
This is of the form:
x2 + y2 + Dx + Ey + F = 0 General Form Where D, E, F are constants
not all zero at a time.
Note: By equation of coefficients:
-2h = D; h = - 1/2 D abscissa of center
-2k = E; k = - 1/2 E ordinate of center
h2 + k 2+ r 2=F ; r =√ (h2 + k 2−F)

Radical Axis of Two Circles

Consider the two non-concentric circles
2 2
x + y −D1 x + E1 y + F 1=0
2 2
x + y −D2 x + E 2 y + F 2=0
The equation:
x 2+ y 2−D1 x + E1 y + F 1+ k ( x 2+ y2 −D2 x + E2 y + F2 ) =0
Represents a circle for any value of k = - 1
If k = 1, the equation of the family of circle above becomes:
( D1−D 2) x + ( E1 −E2 ) y+ ( F1−F 2 )=0
This represents a straight line called the RADICAL AXIS of two circles.

Properties of the Radical Axis

A. If two circles intersect at two distinct points, their radical axis is the common chord of the circles.

Condition for Orthogonality

The two non-concentric circles: x 2+ y 2+ D 1 x + E1 y+ F 1=0
2 2
x + y + D 1 x + E1 y+ F 1=0
Meet at right angles (orthogonal) if: D 1 D 2+ E 1 E 2=2 ( F 1 + F2 )
B. If two circles are tangent, their radical axis is the common tangent to the circles at their point of tangency.

C. The radical axis of two circles is perpendicular to their line centers.

D. All tangents drawn to two circles from a point on their radical axis have equal lengths.

T1 & T2 are points of tangency

PT1 = PT2

Sample Problems:
1. Find the center and radius of circle x2 + y2 + 8x - 10y + 32 = 0
Solution:
Reduce the given equation to standard form by completing the square in x and y:
(x2 + 8x + 16) + (y2 – 10y + 25) = - 32 + 16 +25
(x + 4)2 + (y – 5)2 = 9↔ (x – h)2 + (y – k)2 = r2
Here, C (h, k) = (-4, 5); r = 3

2. Find the equation of the circle passing through the points of intersection of the circles x 2 + y 2 = 16 and
x 2 + y 2 – 10y = 0 and passing through the point (1, 0).
Solution:
The equation of the family is
x 2 + y2 - 25 + k (x 2 + y 2 - 10y) = 0
To determine the particular members of the family,
Solve for k by substituting the coordinates of the point (1, 0),
1+0 – 25 + k (1 + 0 – 0) = 0
k = 24
Substituting k in the equation of the family of circles,
x 2 + y2 - 25 + k (x 2 + y 2 - 10y) = 0
25x2 + 25y 2 - 240y - 25 = 0
Simplifying, 5x2 + 5y2 - 48y – 5 = 0 Answer

3. Find the equation of the radical axis of two circles.

x 2 + y 2 + 10x - 8y + 25 = 0 and x2 + y 2 – 2x + 4y + 1 = 0
Solution:
By getting their difference
x 2 + y 2 + 10x - 8y + 25 = 0
x 2 + y 2 + 2x - 4y + 1 = 0
12x – 12y + 24 = 0
Simplifying, x – y + 2 = 0 radical axis

PARABOLA
The locus of a point that moves in a plane such that its distance from a fixed point equals its distance from a
fixed line.
1. Fixed point is called focus 2. 2. Fixed line is called
directrix
3. Axis - the line passing through the focus and perpendicular to the directrix
4. Vertex - The midpoint of the segment of the axis from the focus to the directrix.
5. Latus rectum - a segment passing through the focus and perpendicular to the axis of the parabola.
6. Focal distance - distance from vertex to focus = a

Standard Equations of Parabola

A. Vertex at V (h, k), Vertical Axis
(x – h)2 = 4a (y – k)
if a is positive (+a) --- concave upward
if a is negative (- a) --- concave downward
Notes:
 1. Equation of axis: x = h
2. Focus: F (h, k + a)
3. End of Latus Rectum
L (h - 2a, k + a)
R (h + 2a, k + a)
4. Equation of Directrix
Y=k-a

B. Vertex at V (h, k), Horizontal Axis

(y – k)2 = 4a (y - k)
if a is positive (+a) --- concave to the right
if a is negative (-a) --- concave to the left directrix
Notes:
1. Equation of the axis: y = k
2. Focus: F (h + a, k)
3. Ends of Latus Rectum:
L (h + a, k + 2a)
axis
R (h + a, k – 2a)
4. Equation of Directrix

C. Vertex at the Origin, Vertex Axis

x2 = 4ay
if a is positive (+a) --- concave upward
if a is negative (- a) --- concave downward
Notes:
1. Axis: the y axis
2. Focus: f (0, a)
3. Latus Rectum: |4 a|
End: L (-2a, a)
R (2a, a)
4. Equation of directrix
y = -a

D. Vertex at the Origin, Horizontal Axis

y2 = 4ax
if a is positive (+a) --- concave to the right
if a is negative (-a) --- concave to the left
Notes:
1. Axis: the x-axis
2. Focus: f (a, 0)
3. Latus Rectum = |4 a|
End: L (a, 2a)
R (a, -2a)
5. Equation of Directrix
x = -a

Remarks:
1. The vertex and focus always lie on the axis of the parabola.
2. Focus is always located on the concave side of the parabola.

General Equations of Parabola

1. Vertical Axis
Ax2 + Dx + Ey + F = 0, E or A must not be zero
2. Horizontal Axis
 Cy2 + Dx + Ey + F = 0, D or C must not be zero

Sample Problems:

1. Find the vertex, focus, equation of the axis, equation of directrix and the ends of the latus rectum of the parabola (x
2
- 2x+1 = 0) and 8y + 15 +1= 0.
Solution:
Since the given equation is linear in y, the parabola has vertical axis.
By completing the square:
(x 2 - 2x + l) = 8y+15 +1=0
(x – 1)2 = 8(y + 2) --- concave upward
a.) Vertex: V (h, k); V (l, - 2)
b.) Focus: F (h, k + a)
Here 4a = 8, a = 2
c.) Equation of Axis:
x = h: x = 1
d.) Equation of Directrix
y = k – a; y = - 4
e.) Ends of Latus Rectum
L (h – 2a, k + a); L (-3, 0)
R (h + 2a, k + a); R (5, 0)

2. A parabolic arc has a height of 20 ft. and a width of 36 ft. at the base. If the vertex of the parabola is at the top of the
arc at, what height above the base is it 18 ft. wide?
Solution:

Determining the equation of the parabola,

In the form x2 = - 4ay (vertex at the origin)
Substitute the point P1 (18, -20) to determine the value of a (18) 2 = - 4a (-20); 4a = 16.2
Now, the equation of the arc is: x2 = - 16.2y
To find the value of y at P (9, y), substitute x = 9
(9) 2 = - 16.2y; y = 5 (distance from the top)
Now, the distance from the base is (20-5) = 15
h = 15 ft. Answer
Short Solution....
By the squared property of parabola
b h1
2

2
= where b = 18 h1 =?
B h2
b = 36 h2 = 20
Substituting

18 h1
2

2
= ; h1 = 5
32 20
Now, h = 20 – h1 = 20 - 5
h = 15 feet Answer

ELLIPSE
Ellipse is the locus of a point P (x, y) in a plane which that the sum of its distances from two fixed points is
constant.
Notes:
1. The two fixed points are called foci
 2. Major axis - the segment cut by the ellipse on the line containing the foci a segment joining the two vertices of
an ellipse of length equal to 12al
3. Diameters - the chords of an ellipse that pass through the center
4. Vertices - the endpoints of the diameter through the foci the endpoints of the major axis
5. Latus Rectum - the segment cut by the ellipse passing through the foci and perpendicular to the major axis
6. Eccentricity - measure the degree of flatness of an ellipse

Standard Equation of Ellipse

Center at C (h, k), Horizontal Major Axis

¿¿
Notes:
1. Major axis: y = k
2. Minor axis: x = h
3. Vertices: V1 (h - a, k)
V2 (h + a, k)
4. Foci: F1 (h - c, k)
F2 (h + c, k)

Center C (h, k), Vertical Major Axis

¿¿
Notes:
1. Major axis: x = h
2. Minor axis: y = k
3. Vertices: V1 (h, k + a)
V2 (h, k – a)
4. Foci: F1 (h, k + c)
F2 (h, k - c)

Center at the Origin, Horizontal Major Axis

¿¿
Notes:
5. Major axis: x = axis
6. Minor axis: y = axis
7. Vertices: V1 (-a, 0)
V2 (a, 0)
8. Foci: F1 (-c, 0)
F2 (c, 0)

Center at the Origin, Vertical Major Axis

¿¿
Notes:
9. Major axis: y = axis
10. Minor axis: x = axis
11. Vertices: V1 (0, a)
V2 (0, -a)
12. Foci: F1 (0, c)
F2 (0, -c)

General Remarks
1. Vertices and foci lie on the major axis
2. |a| is the distance from the center to the vertex
3. |c| is the distance from the center to the foci (focal distance)
4. The ellipse is symmetrical to the major, minor axes and the center

Important Relations
1. a > b, a > c
 2. a2 = b2 + c2
3. e = eccentricity = c/a < 1
4. Latus Rectum, LR = 2b 2 /a

Generation Equation of an Ellipse

Ax2 + Cy2 +Dx + Ey + F = 0 where A ≠ C but of the same sign

Sample Problems

Given the ellipse 16x 2 + 25y 2 – 64x + 50y - 311 = 0

Determine the following
a. Center
b. Length of Major Axis
c. Length of Minor Axis
d. Length of the Latus Rectum b Area
e. Vertices
f. Foci
g. Eccentricity
Solution:
To compute the requirements above the given equation should be transformed into standard form by completing
the square
16(x2 -4x + 4) + 25 (y2 – 2y + 1) = 311 + 16(4) + 25(1)
16(x – 2)2 + 25(y -1)2 = 4
¿¿ ¿¿
Horizontal Major Axis
2
Here a =25 , a=5
2
b =16 , b=4
c =√ ( a −b ) =√ ( 25−16 )=3
2 2 2

a. Center (h, k); C (2, 1)

b. Major axis = 2(a) = 2(5) = 10
c. Minor axis = 2(b) = 2(4) = 8
d. Latus Rectum = 2b 1/a = 2(4) 2/5 = 32/5
e. Vertices: V1 (h – k, k) ; V1 (– 3, 1)
V2 (h + k, k) ; V2 (7, 1)
f. Foci: F1 (h – c, k) ; F1 (– 1, 1)
F2 (h + c, k) ; F2 (5, 1)
g. E = c/a = 3/5
h. Area = πab = π(5)(4) = 20π

HYPERBOLA
Hyperbola is the locus of point P (x, y) in a plane which moves such that the difference of its distances from two
fixed points is a positive constant.

Notes:
1. The two fixed points are called foci
2. Transverse axis - a line segment joining the two vertices of hyperbola
3. The length of the transverse axis is |2a|
4. Conjugate Axis - the perpendicular bisector of the transverse axis
5. The length of the conjugate axis is |2b|
6. Center - point of intersection of transverse and conjugate axis
7. Central Rectangle - the rectangle whose area is (2a) (2b) and whose diagonals are asymptotes of hyperbola
8. Vertices - the endpoints of the transverse axis
9. Asymptotes of the hyperbola - two intersecting lines containing the diagonal of the central rectangle
10. To find the equation of the asymptote, set the right side of the equation of hyperbola in standard form to zero
then solve for y.
11. Central Circle - the circle of radius c with center at the center of the hyperbola circumscribing the central
rectangle
12. Equilateral Hyperbola - hyperbola whose transverse axis equals its conjugate axis
13. Conjugate Hyperbolas - hyperbolas whose transverse axis of one is the conjugate axis of the other
Standard Equations of Hyperbola
A. Center at C (h, k), Horizontal Transverse Axis
¿¿

Notes:

Transverse

1. Transverse axis: y = k axis

2. Conjugate axis: x = h
3. Vertices: V1(h - a, k)
V2(h, k - a)
4. Foci: F1(h - c, k)
F2(h + c, k)

B. Center at C (h, k), Vertical Transverse Axis

¿¿
Notes:
Conjugate
1. Transverse axis: x = h axis
2. Conjugate axis: y = k
3. Vertices: V1(h, k + a)
V2(h, k - a)
4. Foci: F1(h, k + c)
F2(h, k - c)
5. Asymptotes: y - k = ±a/b (x - h)

C. Center at the Origin, Horizontal Transverse Axis

¿¿
Notes:
1. Transverse axis: x - axis
2. Conjugate axis: y - axis
3. Vertices: V1(-a, 0)
4. V2(a, 0)

x
5. Foci: F1(-c, 0)
F2(c, 0)
6. Asymptotes: y = ±b/a x

D. Center at the Origin, Vertical Transverse Axis

¿¿
Notes:
1. Transverse axis: y - axis
2. Conjugate axis: x- axis
3. Vertices: V1(0, a)
V2(0, -a)
4. Foci: F1(0, c)
F2(0, -c)
5. Asymptotes: y = ±a/b x
 General Remarks
1. Vertices and foci are on the transverse axis
2. |a| is the distance from the center to the vertex
3. |c| is the distance from the center to the focus
4. The hyperbola is symmetrical to the transverse and conjugate axis and to the center.

General Relations
l. c > a, c > b (a = b or a < b or a > b)
If a = b, then the hyperbola is called equilateral hyperbola
2. c2 = a2 + b2
3. Length of Latus Rectum = 2b2 /a
4. Eccentricity, e = c/a > 1

General Equation of Hyperbola

Ax2 + Cy2 + Dx + Ey + F = 0 where A and C are of opposite signs

Sample Problem:
1. Find the equation of the hyperbola, center (2,0) focus (2,3) eccentricity = √ 3
Solution:
Since the center and focus have the same abscissas the axis of hyperbola is vertical
c
e= = √3 ; √ 3 a
a
C=3–0=3 focal distances
Solving for a:
From c 2 = a2 + b2 ; b 2=c 2−a 2
2 2
b =c −¿
2
b =9−3=6
Equation: ¿¿
¿¿

Simplifying, 2y2 – x2 + 4x – 10 = 0 Answer

THE CONIC SECTIONS (a summary)

A conic section is the locus of a point which moves such that its distance from a fixed point called focus is
in constant ratio and called eccentricity to its distance from a fixed straight line directrix.

A. General Form of a Quadratic Equation in x and y

Ax2 + Cy2 + Dx + Ey + F = 0
1. Ellipse : A ≠ C, same signs
2. Circle : A = C, same signs
3. Hyperbola : A and C have opposite signs
B. Eccentricity e = c/a
1. Circle :e=0
2. Parabola :e=1
3. Ellipse :e<1
4. Hyperbola : e > 1
Notes;
1. The circle, parabola, ellipse and hyperbola are called conic sections (or conics) because any one of them can be
obtained geometry by cutting a cone with a plane.
 Ellipse Circle Parabola Hyperbola

2. If the Cutting plane is perpendicular to the axis of the cone, the section is a circle.
3. If the cutting plane is making an angle (other than 90 0) with the axis of the cone, the section is an eclipse
4. If the cutting plane is parallel to one of the elements of a cone, the section is a parabola
5. If the cutting plane is parallel (but not coincident) to the axis of the cone, the section is a hyperbola.

In each of the cases, the cutting plane should not pass through the vertex of the cone, otherwise the section that will be
formed is a degenerate conic.

Degenerate Conic (one point, one line, two lines) is a conic formed if the cutting plane is passing through the vertex
or along one of its elements
Principal Axis of a Conic - is the line through the focus and perpendicular to the directrix
Diameter of a Conic - the locus of the midpoints of a system of parallel chords.

ROTATION OF AXES
Let the x-axis be rotated at an angle θ and let α be the angle formed by Ox' and OP
Consider:

From the illustration above:

x ' = r cosα and y’ = r sin α
Similarly,
x = r cos (θ + α )
x = r cosθ – r sin α – r sin θ sin α
y = r sin (θ + α )
y = r sinθ – r cos α – r cos θ sin α
Substituting equation in 1 in x and y above,

x = x' cosθ - y' sin

θ y = x' sinθ + y' Previous coordinates in terms of new coordinates
cosθ

Solving for x' and y'

x' = x cosθ - y sin θ
y' = - x sinθ + y cos New coordinates in terms of previous coordinates
θ

Sample Problem;
That is the point (2, 3) in the new x' - y' system if the x- axis is rotated at an angle 30 o?
Solution:
 x = 2, y = 3, θ = 30o
x' = x cos θ + y sin θ = 2 cos 30o + 3 sin 30o = 3.232
y' = - x sinθ + y cosθ = -2 sin 30o + 3 cos θ 30o = 1.598
P' = (3. 232, 1.598) Answer

GENERAL SECOND-DEGREE EQUATION in x and y

Ax2 + By2 + Cy2 + Dx + Ey + F = 0, B ≠ 0

Notes
1. The equation above represents a parabola, an ellipse, a hyperbola, two straight lines, a point or no locus at all.
2. The cross-product term (xy) can be eliminated upon rotation of axes by an amount of angle θ , expressed as,

A−C
cot 2θ =
B

Sample Problem:
Find the angle of rotation that will eliminate the xy term in the equation 2x 2 + xy + 2y 2 = 1
Solution:
A = 2, B = 1, C = 2
A−C 2−2
cot 2 θ= = =0
B 1
−1
2 θ=cot 0=90 °
θ=45 ° Answer

INVARIANTS AND DISCRIMINANT

For any rotation of axes, the coefficients A, B, C and A', B', C' satisfy the equation
B2 – 4AC = B'2 - 4 A' C'
Notes:
1. The quantity B2 - 4AC is called invariant under a rotation of axes
2. If the rotation yields B' = 0, the right side is just -4A'C' and B2 – 4AC is expressed as discriminant
Nature of a conic if:
l. B2 – 4AC = 0, parabola
2. B2 – 4AC < 0, ellipse
3. B2 – 4AC > 0, hyperbola

TRANSLATION OF AXES
Let the origin be shifted to new location at O' (h, k), so that x' and y' are the new coordinate.

Consider

Relations:

x = x' + h or x' = x
–h
y = y' + h or y' = x
–k

Sample Problems:
1. What is the point (4, 5) in the new system if the origin is shifted to the point (-1, 3)
Solution:
x = 4, y = 5, h = 1, k = 3
 x' = x – h = 4 – (-1) = 5
y' = y – k = 5 – (-3) = 2
Thus, P' (5, 2) Answer
2. That is the equation 3x - 2y = 5 in the new system if the origin is shifted to the points (5, -1)
Solution:
h = 5, k = 1
x = x' + h = x' + 5
y' = y + k = y' - 1
Substitute x and y in the equation given above
3 (x' + 5) – 2 (y' – 1) = 5
Simplifying
3x' - 2y' = -12 Answer
SUPPLEMENTARY PROBLEMS
1. A point P (x, 3) is equidistant from the points A (l, 5) and B (-l, 2). Find x. Ans. 3/4
2. Find the locus of points P (x, y) such that the distance from P to (3, 0) is twice its distance to (1, 0).
Ans. 3x2 – 3y2 – 2x – 5 = 0
3. Find the length of the segment joining the two midpoints of the sides of the triangle if the length of the third side
opposite to it is 30 cm. Ans. 15 cm
4. A line from, (1, 4) to Q (4, -1) is extended to a point R so that PR = 4PQ. Find the coordinate R. Ans. R (13, -16)
5. Two vertices of a triangle are (0, -8) and (6, 0). If the medians intersect at (9, -3), find the third vertex of the triangle.
Ans. (- 3, - 1)
6. The area of a triangle with vertices (6, 2), (x, 4) and (0, - 4) is 26. Find x. Ans. -2/3 and 50/3
7. Find the length of the median from A of a triangle ABC given vertices A (1, 6), B (- 1, 3) and C (3, -3). Ans. 6
8. If the midpoint of a segment is (5, 2) and one endpoint is (7, -3), what are the coordinates of the other end? Ans. (3, 7)
9. Given vertices of a triangle ABC:
A (l, 5), B (-1, 1) and C (6, 3). Find the intersection of the median. Ans. (2, 3)
10. Find the inclination of the line 2x + 5y = 10. Ans. 158.2o

EQUATION OF A STRAIGHT-LINE SUPPLEMENTARY PROBLEMS

Find the equations of the line/s satisfying the given condition
1. Passing through (1, -2) and perpendicular to the line through (2, - 1) and (- 3, 2)
Ans. 5x - 3y - 11 = 0
2. With x intercept of 5 and passing through (3, 4). Ans. 2x + y - 10 = 0
3. Passing through (-3, 4) and with equal intercepts. Ans. x - y + 7 = 0, x + y – 1 = 0
4. Making an angle of 45o with axis and passing through (2, 3) Ans. x - y + 1 = 0
5. With slope -12/5 crosses the first quadrant and forms with the axes a triangle with perimeter of 15. Ans. 12x + 5y =
30
6. Passing through (7, - 4) and at a distance of 1 unit from the point (2, 1)
Ans. 4x + 3y – 16 = 0 ; 3x + 4y – 5 = 0
7. Passing through the midpoint of the segment joining the points (1, 3) and (5, 1) and parallel to the line 3x – y + 5 = 0.
Ans. 2x – y – 4 = 0
8. Find the value of parameter k so that the line 3x - 5ky + 5 = 0
a) will pass through (0, 1)
b) will be parallel to x + 2y = 5
c) will be perpendicular to 4x + 3y = 2
d) has the y - intercept equal to 3
Ans. a) 1 b) 8/5 c) 4/5 d) 1/3
9. Find the equations of the lines parallel to the line x + 2y - 5 = 0 and passing at a distance 2
from the origin. Ans. x + 2y + 2√ 5 = 0 and x + 2y - 2√ 5 = 0
10. Find the equation of the perpendicular bisector of the segment joining (2, 5) and (4, 3).
Ans. x – y + 1 = 0
11. Given vertices of a triangle ABC, A (2, 0), B (3, -2) and C (7, 5)
a) Find the equation of the median from A. Ans. a) x – 2y – 2 = 0
b) Find the equation of the altitude from B. Ans. b) x + y – 1 = 0
c) The intersection of medians from B to C Ans. c) (4, 1)
12. Find the normal intercept and the normal angle of line 5x + 12y - 39 = 0
Ans. ρ=3 ,θ = 67. 380

CIRCLES SUPPLEMENTARY PROBLEMS

1. Find the center and radius of the circle whose equation is x2 + y2 - 4x - 6y - 12 = 0
(ECE Board Problem -Oct 1981) Ans. C (2, 3) r = 5
2. Find (Board Problem - Mar. 1981) Ans. 25 π sq. units
Find the equation of the circle whose center is at (3, 5) and whose radius is 4 units.
Ans. (x – 3)2 + (y + 5)2 = 1
For Problems 4 - 9 determine the equation of the circle given the following conditions
3. Passes through the point (2, 3), (6, 1) and (4, -3) Ans. x2 + y2 - 6x - 1 = 0
4. Center on the y - axis and passes through the origin and point (4, 2).
Ans. x2 + y2 - 10 = 0
5. Passes through the points of intersection of the circles x2 + y2 = 5, x2 + y2 - x + y = 4, and the point (2, -3)
Ans. x2 + y2 - 2x + 2y – 3 = 0
6. Center on the line x - 2y – 9 = 0 and passes through the points (7, -2) and (5, 0)
Ans. x2 + y2 - 10x + 4y + 25 = 0
7. Circumscribing the triangle determined by x - y- 8 = 0, x = - y and y = -1
Ans. x2 + y2 - 8x + 2y +8 = 0
8. Given the endpoints of the diameter (5, 2) (-1, 2) Ans. x2 + y2 - 4x - 4y - 1 = 0
9. Find the equation of the line tangent to the circle x2 + y2 - 8x - 8y + 7 = 0 at the point (1, 0)
Ans. 3x + 4y – 3 = 0
PARABOLA SUPPLEMETARY PROBLEMS
1. Find the vertex, focus, and end points of the Latus Rectum each of the following parabolas

( −14 ,− 14 ) , F ( −316 ,− 14 ), EL ( −316 ,− 14 ± 18 )

a . 4 y 2−x+ 2 y =0 Ans . V

b .2 y −5 x+ 3 y−7=0 Ans . V ( ,− ) , F (−1 ,− ) , EL(−1 ,− , ± )

2 −13 3 3 3 5
8 4 4 4 8
2. Find the equation of the parabola determined by the given conditions
a. focus at (-11/4, 1) and the endpoint of latus rectum is (-11/4, 5/2) Ans: y2 + 3x + 2y + 7 = 0
b. vertex at (1, -1) and focus at (1, - 3/4) Ans: x2 - 2x – y = 0
c. vertex at (0, 3) directrix x = -1 Ans: y2 - 4x - 6y + 9 = 0
3. axis vertical, vertex (-1, -1) and passing through (2, 2) Ans: x2 - 2x - y = 0
4. A chord passing through the focus of the parabola y2 = 16x has one end at the point (1, 4). Where is the other end of
the chord? Ans: (4, 8)
5. Find the equation of the line tangent to the parabola x2 + 2x + 3y - 1 = 0 Ans: 2x + 3y - 1 = 0
6. Find the equation of the circle that passes through the vertex and the endpoints of the latus rectum of the parabola y 2 =
8x.
Ans: x2 + y2 - 10x = 0
7. Find the equation of parabola whose axis is horizontal, vertex is on the y - axis and which through (2, 4) and (8, - 2)
Ans: y2 = 20y - 18 + 100 = 0
y2 = 4y + 2x – 4 = 0
8. An arch in the form of parabolic curve, with a vertical axis is 60m, across the bottom. The highest point is 16m above
the horizontal base. What is the length of a beam placed horizontally across the arch 3m below the top'? Ans: 26m
9. Assume that water issuing from the end of a horizontal pipe, 25 ft above the ground described a parabolic curve, the
vertex of the parabola being at the end of the pipe, the flow of water has curve outward 10 ft. beyond a vertical line
through the end pipe, how far beyond this vertical line will the water strikes the ground? Ans: 17.68 ft.

ELLIPSE SUPPLEMETARY

1. In the ellipse below determine the following: a) Center, b) Vertex, c) Foci, d) Major Axis, e)
Minor Axis, f) Latus Rectum, g) Eccentricity

A. 25x2 + 16y2 – 50x + 32y - 1559 = 0

B. 144 x2 + 169y2 + 864x - 23,760 = 0
Ans.
A. a) C (1, -1); b) V1 (1, 9), V2 (1, -11); c) F1 (l, 5), F2(1, -7); d) 20; e) 16; f)12.8; g) 0.6
B. a) C (3, 0); b) V1 (- 16, 0), V2(10, 0); c) F1 (- 8, 0), F2(2, 0); d) 26; e) 24; f) 205/13; g) 6/13

2. In each of the following find the equation of ellipse satisfying the given conditions
A. center at (0, 0), focus at (±√ 3, 0), and b: 1 Ans: x2 + 4y2 = 4
B. center at (1, 0), focus at (1, √ 3) e: √ 3/2 Ans: 4y2 + y2 = 8x
Focus at (0, -1), (-4, -1), a = √ 6 Ans: x2 + 3y2 + 4x + 6y + 1 = 0
C. center at (-1/2, 2) a = 5/2 b= 2 Ans: 16x2 + 25y2 + 16x + 4 = 100y
D. center at (0, 0), vertex (0, 4) e = 1/2 Ans: 4x2 + 3y2 – 48 = 0
3. A satellite orbits around the earth in an ellipse orbit of eccentrically of 0.80 and semi - major axis length 20, 000 km.
If the center of the earth is at one focus, find the maximum altitude (apogee) of the satellite. Ans: 36, 000 km
4. Find the equation of the locus of a point which moves so that the sum of its distance from (-2, 2) and (1, 2) is 5.
Ans: 16 x2 + 25y2 + 16x – 100y + 4 = 0

5. Find the eccentricity of an ellipse whose major axis is thrice a long as its minor axis Ans:2
√ 2
3
6. That is the quadrilateral formed by joining the foci of an ellipse to the endpoints of the minor axis? Ans: rhombus
7. Find the distance of the point (3, 4) to the foci of the ellipse whose equation 4x 2 + 9y 2 = 36 Ans: 3 ± √ 5
 8. An arch in the form of a semi - ellipse has a span at 45 m and its greatest height is 12m. There are two vertical
supports equidistant from each other and the ends of the arc. Find the height of the support. Ans: 8 √ 2m.
9. Determine the locus of a point P (x, y) so that the product of the slopes joining P (x, y) to (3, -2) and (-2, 1) is -6.
Ans: 6x2 + y2 - 6x + y - 20 = 0
10. What is the area of the ellipse whose equation is 25x 2 + 16y 2 = 400 Ans. 20π

HYPERBOLA SUPPLEMENTARY PROBLEMS

1. Find the center, vertices, foci and asymptotes, transverse axis, conjugate axis, latus rectum and eccentricity of
hyperbola below.
a. 9x2 - 16y2 + 18x + 64y = 91
b. 16x2 - 4y2 – 62x + 24y + 92 = 0
c. 25x2 - 16y2 = 400

a) C (-1, 2); V (-1∓2, 2); F (-1∓5/2, 2); asymptote (3x + 4y – 5 = 0, 3x – 4y + 11 =0); TA = 4; CA = 3; LR

= ¾; e = 5/4
b) C (2, 3); V1 (2, 1), V2 (2, 7); F1 (2,3+√ 20), F2 (2,3 - √ 20); asymptote (2x - y – 1 = 0, 2x + y + 7 =0); TA
= 8; CA = 4; LR = 2; e = √ 6/2
c) C (0, 0); V1 (-4, 0), V2 (4, 0); F1 (-√ 41, 0), F2 (√ 41 , 0); asymptote (y = ∓( 5/4x); TA = 8; CA = 10; LR =
25/2; e = √ 41/4

2. Find the equation of the hyperbola satisfying the conditions given in each case
a. Center (3, -1); vertex (l, -1); focus (0, 1) Ans: 5x2 - 4y2 - 30x - 8y + 21=0
b. vertices at (0, 4) and (4, 4); foci at (5, 4) and (-1, 4)
Ans. 5x2 -4y2 + 32y- 64 = 0
c. Center at (1, 1), vertex (l, 3), eccentricity =2 Ans: x2 - 3y2 - + 10 = 0
d. Directrices: y = ± 4 ; asymptote: y = ± 3/2 x Ans: 9y2 - 324x2 - 208 = 0
e. Asymptotes: 3y=± 4 ; foci (± 6, 0) Ans: 400 x2- 225y2 – 5184 = 0
f. Foci (0, 0), (0, 10); asymptote: x+y=5 Ans: 2x2 - 2y2 + 20y - 25 = 0
g. Asymptote: x+y=1 and x-y=1 and passing through (-3,4) and (5, 6)
Ans. x2- y2-2x – 2 = 0
h. Axes along the coordinate’s axes, passing through (-2 5) Ans: 4y2 - 5x2 = 19
i. Vertices at (0, ±4) passing through (-2, 5) Ans: 4y2 - 9x2 = 64
3. Find the eccentricity of a hyperbola whose transverse axis and conjugate axis are equal in length Ans: e=√ 2

ROTATION OF AXES SUPPLEMENTARY PROBLEMS

1. What is the point (3, 4) in the new x' - y' system if the axes is rotated at an angle of 40 ° ?
Ans. P' (4.869, l.136)
2. What is the equation x 2+ xy + y 2 = 1 in the new coordinate system if the x-axis is rotated to eliminate the constant
term? Determine the type of conic section formed.
Ans. 3(x’¿2 + (y’¿2 =2 or (x’¿2 + 3(y’¿2 , ellipses
3. Using the discriminant, classify each of the following as circle, ellipse, parabola or hyperbola.
a. x 2−xy + y 2 + x− y=5 Ans . ellipse
b. 3 x 2+ 6 xy + 3 y 2 −4 x +7 y=12 Ans . parabola
c. 2 x 2+ 4 xy− y 2 −2 x +5 y=6 Ans . hyperbola
TRANSLATION OF AXES

SUPPLEMENTARY PROBLEMS

1. Find the point (5, - 3) in the new system if the origin is shifted to the point (-1, 2). Ans. P'(6, -5)
2. The new point is (6, -2) after the origin is shifted to the point (3, - 2). What is the previous point? Ans. P (9, - 4)
3. What is the circle x 2 + y 2 – 2x - 3 = 0 in the new system if the origin is shifted to (1, 2)? Ans. (x’¿2 + (y’¿2 +4y’ = 0

REVIEW PROBLEMS
1. The Y-axis is also called____.
A. ordinate C. point
B. abscissa D. angle

2. In polar coordinate system, the distance from a point to the pole is known as:
A. Polar angle C x-coordinate
B. Radius vector D. y-coordinate

3.The distance of a point from the y-axis is called___.

A. abscissa C. x-intercept
B. y-intercept D. ordinate

4. The distance of a point from the x-axis is called___.

A. abscissa C. x-intercept
B. y-intercept D. ordinate

5. State the quadrant in which the coordinate (15, - 2) lies

A. III C. I
B. II D. IV

6.The product of the slopes of the equation of 2 lines is -1. One of the lines is___.
A. Non-intersecting C. Perpendicular
B. Parallel D. Skew

7. The equation of a line is y= mx + b, what is m?

A. x-intercept C. y-intercept
B. Slope D. eccentricity

8. Find the equation of a line with slope 3 and y-intercept -2.

A. y=2x+3 C. y= -3x+2
B. y= 2x-3 D. y=3x-2

9. Find the point on the line 3x + y +4 = 0 that is equidistant from the points (-5, 6) and (3, 2).
A. (-2,2) C. (-2, -2)
B. (-2,3) D. (2,2)

10. Determine the equation of the line passing through the point (1, 17) and (13, 4).
A. 13x - 12y - 217 = 0 C. 13x+12y – 217 = 0
B. 13x - 12y + 217 = 0 D. 13x+ 12y + 217 = 0

11. Find the equation of the line that passes 3 units from origin and parallel to 3x – 4y + 5 = 0
A. 3x – 4y + 15 = 0 C. 3x – 4y - 8 = 0
B. 3x – 4y + 5 = 0 D. 3x – 4y + 8 = 0

12. Find the inclination of the line 2x + 5y = 10

A. 140.236° C. 158.2°
B. 130.57° D. 157.3°

13. What is the acute angle between the lines y = 3x + 2 and y = 4x + 9?

A.4.4 deg C. 28.3 deg
B. 5.2 deg D. 18.6 deg

14. Find the area of the triangle formed if the sides are given by the equations: 2x - 3y + 21 = 0, 3x-2y-6 = 0, 2x+3y+9 = 0
A. 75 C. 86
B. 97.5 D. 28/3

15. Find the area of the triangle having the vertices at -4 - i, 1 + 2i, 4-3i.
A. 7 C.10
B. 17 D. 107
16. Find the equation of the line passing 3 units from the origin and parallel to 3x + 4y – 10 = 0
A. 3x + 4y – 5 = 0 C. x - 3y + 15 = 0
B. 4x + 3y + 1 = 0 D. 3x + 4y – 15 = 0

17. Find k in the equation of the line 5x – 2y + k = 0 that is tangent to y = 6 + x 2.

A. 25/8 C. 23/4
B. 5/12 D. 71/8

18. What is the equation of the line that passes through (- 3, 5) and is parallel to the line 4x - 2y+2 = 0?
A. 4x-2y + 12 = 0 C. 4x + 2y -11 = 0
B. 2x + y + 10 = 0 D. 2x-y + 11 = 0

19. Find the equation of the circle whose center is at (3, -5) and whose radius is 4.
A. x2 + y2 -6x + 10y +18 = 0 C. x2 + y2 - 6x - 10y +18 = 0
B. x2 + y2 +6x + 10y +18 = 0 D. x2 + y2 + 6x - 10y +18 = 0

20. Find the radius of the circle x2 + y2 - 4x + 8y = 7.

A. √ 17 C. 3√ 5
B. 3√ 3 D. 5

21. Find the coordinates of the center of a curve whose equation is x2 + y2 - 4x - 6y – 12 = 0.

A. C (2,-3) C. C (2,-3)
B. C (2,3) D. C (-2,-3)

22.What is the equation tangent to the curve x2 + y2 = 5 at (2,1)

A. 2x + y = 25 C. 2x + y = 5
B. x - 2y = 0 D. x + 2y = 5

23. A circle with center at the origin has a radius of 5. Find the equation of a parabola opening to the right that has its
vertex on the circle and crossing the points of intersection of the circle and y-axis.
A. 5x + 25 = y2 C. 5y + 55 = x2
B. 5x - 25 = y2 D. 5y - 55 = x2

24. What conic section is having an equation x2 - 4y + 3x -8 = 0

A. Parabola C. Hyperbola
B. Ellipse D. circle

25. What is the coordinate of the vertex of the equation x2 + 2x - y + 2 = 0.

A. origin C. (-1, -1)
B. (-1, 1) D. (1, -1)

26. An arch is in the form of an inverted parabola and has span of 12 feet at the base and a height of 12 feet. Determine
the equation of parabola and give the vertical clearance 4 feet from the vertical centerline.
A. 7.33 ft C. 5.33 ft
B. 6.00 ft D. 6.67 ft

27. A parabola opening to the right that has its vertex on the circle and crosses the circle of unit radius thru the diameter.
If the center of the circle is at the origin, find the coordinates of the focus of parabola.
A. (1/3, 0) C. (-1/3, 0)
B. (3/4, 0) D. (-3/4, 0)

28. Find the family of parabola with vertex at the origin and focus at the x-axis.
A. Cy = x2 C. Cx = y2
A. y = x + C D. x = y2 + C
2

29. What is the equation of a circle that passes thru the vertex and latus rectum of the
curve y2 = 8x?
A. (x- 4¿2 + y 2 = 20 C. (x- 2¿2 + y 2 = 16
B. (x- 5¿2 + y 2 = 25 D. (x- 3¿2 + y 2 = 17

30. What kind of a curve is 4x2 - 6y2 + 5y + 6x +5 = 0.

A. Circle C. Parabola
B. Ellipse D. Hyperbola

31. Axis of hyperbola through its foci.

A. Latus Rectum C. Transverse Axis
B. Directrix D. Major Axis
32. Find the center of the curve 2y2 - x2 -4x - 10 = 0.
A. C (2,0) C. C (-2,0)
B. C (2,1) D. C (1,2)

33. Find the equation of the hyperbola whose asymptotes are y = ± 2x and which passes through (5/2, 3).
A. 3x2 - y2 = 9 C. 4x2 - y2 = 16
B. 5x - y = 25 D. 2x2 - y2 = 4
2 2

34. Find the eccentricity of the curve 9x2 - 4y2 - 36x + 8y = 4.

A. 1.80 C. 1.92
B. 1.86 D. 1.76

35. Find the length of the latus rectum of 9x2 - 4y2 - 36x + 16y = 0
A. 5.433 units C. 9.218 units
B. 6.708 units D. 8.694 units

36. Determine the equation describing the locus of point P (x, y), such that the sum of the distances between P and (- 5, 0)
and between P and (5, 0) is constant at 20 units.
A. (x/10)2 + (y/8.66) 2 = l C. (x/5)2 + (y/10) 2 =1
2 2
B. (x/10) + (y/5) = l D. (x/8.66) 2 + (y/10)2 = 1

37. Given the curve 16x2 + 25y2 - 64x - 50y – 311= 0, determine its center.
A. (1,2) C. (2, -1)
B. (2,1) D. (1, -2)

38. A curve has the equation 16x2 + 9y2 - 32x - 128 = 0. Its eccentricity is:
A. 0.88 C. 0.77
B. 0.66 D. 0.55

39. Given the equation of the curve 9x2 +25y2-144x +200y + 751 = 0. Find the distance between the foci.
A. 6 C. 10
B. 8 D. 12

40. What conic section, B2- 4AC = 0?

A. Circle C. Parabola
B. Ellipse D. Hyperbola

41. What conic sections, B2- 4AC > 0?

A. Circle C. Parabola
B. Ellipse D. Hyperbola

42. When discriminant of a conic section, D = B2 - 4AC = 1, the curve is a __.

A. Ellipse C. parabola
B. circle D. hyperbola

43. That is the eccentricity, e of a conic section with discriminant B 2 - 4AC = 0.

A. e = 0 C. e > 1
B. e < 1 D. e = l

44. The eccentricity of a given curve is equal to zero, the given curve is:
A. parabola C. hyperbola
B. ellipse D. circle

45. In a conic section if the discriminant D = 1, then the locus is

A. hyperbola C. circle
B. parabola D. ellipse