Analytic Geometry _2012

PART I – CONTENT UPDATE

Philippine Normal UniversityLICENSURE EXAMINATION FOR TEACHERS (LET)

Refresher Course

Introduction

This material includes a brief review of the basic terms concerning lines, circles, parabolas, ellipses, and hyperbolas.

A straight line is represented by an equation of the first degree in one or two variables, while the circle, parabola, ellipse and hyperbola are represented by equations of the second degree in two variables.

A. STRAIGHT LINE

Figure 1

Analytic GeometryRosemarievic V. Diaz, Ph.D.

WHAT TO EXPECTMAJORSHIP: MATHEMATICS

FOCUS: Analytic Geometry

LET COMPETENCIES:

1. Determine the equation of a line given: a. any two points on the lineb. a point and the slope of the linec. a point and the slope of the line parallel to the desired lined. a point and the slope of the line perpendicular to the desired linee. the intercepts.

2. Solve problems involving a. the midpoint of a line segment, distance between two points, slopes of lines,

distance between a point and a line, and segment division.b. a circle, parabola, ellipse, and hyperbola.

3. Determine the equations and graphs of a circle, parabola, ellipse and hyperbola.

Prepared by: ROSEMARIEVIC V. DIAZ

1. Distance between two points

The distance between two points P (x1 , y1) and Q (x2 , y2) is

PQ=√ (x2−x1 )2+( y2− y1)2=√ (x1−x2 )2+( y1− y2)

2.

Example 1: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the length of AB .

Solution: AB=√ [1−(−2 ) ]2+ [ (−5 )−4 ]2=√ [ (−2 )−1 ]2+ [4−(−5 ) ]2=√90=3√10Hence, AB=3√10 units.

2. Slope of a line

The slope m of the non-vertical line containing P (x1 , y1) and Q (x2 , y2) is

Example 2: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the slope of AB .

Solution:m=

(−5 )−41−(−2 )

=4−(−5 )(−2 )−1

=−3

Hence, the slope of AB is −3 and the line leans to the left.

Example 3: Given the points X (0 , 9 ) , Y (5 ,−4 ) , and Z (0 , −4 ) . Determine the slopes

of XY , YZ , and ZX . Sketch triangle XYZ .

Solution: slope of XY=

9−(−4 )0−5

=−135 . Note that XY leans to the left.

slope of YZ=−4−(−4 )

5−0=0

. Note that YZ is a horizontal line.

slope of ZX=−4−9

0−0 is undefined. Thus, ZX is a vertical line.


Note:a. The slope of a line parallel to the x -axis is 0.b. The slope of a line parallel to the y -axis is undefined.c. The slope of a line that leans to the right is positive.d. The slope of a line that leans to the left is negative.

Sketch:

Figure 2

3. The equation of a line

In general, a line has an equation of the form ax+by+c=0 , where a , b , and c are real numbers and that a and b are not both zero.

Example 4: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the equation of A⃗B .

Solution: Using point A (1 ,−5 ) , the equation is

y− (−5 )= (−5 )−41−(−2 )

( x−1 )

y+5=−3 ( x−1 )y+5=−3 x+33 x+ y+5−3=03 x+ y+2=0

Hence, the equation of A⃗B in general form is 3 x+ y+2=0 .

4. Different forms of the equation of a line

a. General form: ax+by+c=0 .

b. Slope-intercept form: y=mx+b , where m is the slope and b the y -intercept.

c. Point slope form: y− y1=m (x−x1) , where (x1 , y1) is any point on the line.

d. Two point form: y− y1=

y2− y1x2−x1

(x−x1 ), where (x1 , y1) and (x2 , y2 ) are any two

points on the line.

e. Intercept form:

xa+ yb=1

where a is the x -intercept and b the y -intercept.

5. Parallel and perpendicular lines

Given two non-vertical lines g and s so that g has slope m1 and s has slope m2 .

a. If g and s are parallel, then m1=m2 .


Z

X

Y

b. If g and s are perpendicular, then m1=

−1m2 .

Example 5: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the equation of the

line passing through the point J (8 ,−7 ) and parallel to A⃗B .

Solution: The slope of A⃗B=

(−5 )−41−(−2 )

=−3.

The equation of the line through J and parallel to A⃗B is computed as follows:

y− (−7 )=−3 ( x−8 )

Thus, is the equation of the line passing through J (8 ,−7 ) and parallel to .

Example 6: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the equation of the

line passing through the point J (8 ,−7 ) and perpendicular to A⃗B .

Solution: The slope of the line perpendicular to A⃗B is

13 .

The equation of the line passing through J and perpendicular to A⃗B is computed as follows:

y− (−7 )= 13

( x−8 )

3 ( y+7 )=x−83 y+21=x−8−x+3 y+21+8=0−x+3 y+29=0

or x−3 y−29=0

Thus, x−3 y−29=0 is the equation of the line passing through J (8 ,−7 ) and perpendicular to A⃗B .

6. Segment division

Given segment PQ with P (x1 , y1) and Q (x2 , y2) :

a. the midpoint M of segment PQ is M ( x1+x22

,y1+ y22 )

.


Figure 3

b. If a point D divides PQ in the ratio

r1r2 so that

PDDQ

=r1r2 , then

P( r1 x2+r2 x1r1+r2,r 1 y2+r2 y1r 1+r2 )

.

Figure 4

Example 7: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the coordinates of M , the midpoint of segment AB .

Solution:M ( 1+ (−2 )

2, (−5 )+4

2 ). Thus, the midpoint is

(−12 , − 12 ).

Example 8: Given the points A (1 ,−5 ) and B (−2 , 4 ) . Determine the coordinates of

so that the segment AB is divided in the ratio 3 :5 .

Solution: In this problem, either

ACCB

=35 or

BCCA

=35 . Consequently, there are two

possible answers.

Case 1: Let

ACCB

=35 , then

C ( (3 ) (−2 )+ (5 ) (1 )3+5

, (3 ) (4 )+(5 ) (−5 )3+5 )

.

Hence, the point of division is C (− 18 , −138 )

.

Case 2: Let

BCCA

=35 , then

C ( (3 ) (1 )+(5 ) (1−2 )3+5

, (3 ) (−5 )+ (5 ) (4 )3+5 )

.

Hence, the point of division is C (−78 , 58 ) .

7. Distance of a point from a line


The distance d of a point P (x1 , y1) from the line Ax+By+C=0 is given by

d=|Ax1+By1+C|

√A2+B2 .

Figure 5

Example 9: Determine the distance of Z (−2 , 5 ) from the line 2 x−3 y+8=0 .

Solution:

d=|(2 ) (−2 )+(−3 ) (5 )+8|

√(2 )2+(−3 )2=11

√13 or

11√1313

B. CIRCLE

1. Definition

A circle is a set of all points on a plane that are equidistant from a fixed point on the plane. The fixed point is called the center, the distance from the center to any point of the circle is called the radius.

Figure 6

2. Equation of a circle

a. General form: x2+ y2+Dx+Ey+F=0

b. Center-radius form: ( x−h )2+( y−k )2=r2 where the center is at (h , k ) and the radius is equal to r .


Figure 7

Example 10: What is the general form of the equation of a circle with center (−3 , 1 ) with radius 3?

Solution: [ x−(−3 ) ]2+( y−1 )2=32

( x+3 )2+( y−1 )2=9x2+6 x+9+ y2−2 y+1=9x2+ y2+6 x−2 y+1=0 is the equation of the given circle.

3. Line tangent to a circle

A line tangent to a circle touches the circle at exactly one point called the point of tangency. The tangent line is perpendicular to the radius of the circle, at the point of tangency.

Example 11: A circle has its center C (3 , −1 ) . Supposed that F (2 , 6 ) is on the circle. What is the equation of the line tangent to the circle C and passes thought the point F .

Solution: The slope of CF is

6− (−1 )2−3

=−7.

The slope of the tangent line (which is perpendicular to CF ) is

17 .

The equation of the tangent line at F is

y−6= 17

(x−2 )

7 y−42=x−2−x+7 y−40=0 or x−7 y+40=0

C. CONIC SECTION

A conic section or simple conic, is defined as the graph of a second-degree equation in x and y .

In terms of locus of points, a conic is defined as the path of a point, which moves so that its distance from a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus of the conic, the fixed line is called the directrix of the conic, and the constant ratio is called the eccentricity, usually denoted by e .

If e<1 , the conic is an ellipse. (Note that a circle has e=0 .)If e=1 , the conic is a parabola.


If e>1 , the conic is a hyperbola.

Figure 8

D. PARABOLA

1. Definition

A parabola is the set of all points on a plane that are equidistant from a fixed point and a fixed line of the plane. The fixed point is called the focus and the fixed line is the directrix.

Figure 9

2. Equation and graph of a parabola

a. The equation of a parabola with vertex at the origin and focus at (a , 0 ) is y2=4 ax .

The parabola opens to the right if a>0 and opens to the left if a<0 .

b. The equation of a parabola with vertex at the origin and focus at (0 , a ) is x2=4 ay .

The parabola opens upward if a>0 and opens downward if a<0 .

c. The equation of a parabola with vertex at (h , k ) and focus at (h+a , k ) is

( y−k )2=4 a (x−h ) . The parabola opens to the right if a>0 and opens to the left if a<0 .

d. The equation of a parabola with vertex at (h , k ) and focus at (h , k+a ) is

( x−h )2=4a ( y−k ) . The parabola opens upward if a>0 and opens downward if a<0 .

e. Standard form: ( y−k )2=4 a (x−h ) or ( x−h )2=4a ( y−k )

f. General form: y2+Dx+Ey+F=0 or x

2+Dx+Ey+F=0

3. Parts of a parabola


L1

L2

M

y

xO

C1

C2

F(a, 0)

P(x, y)

F(0,-4/3)P(x,y)

O

y

x

y – 4/3 = 0

O

y

x

L1

L2

V(3,2)F(5,2)

x 1 = 0

y 2 = 0

yx3162 382 2 xy

a. The vertex is the point, midway between the focus and the directrix.b. The axis of the parabola is the line containing the focus and perpendicular to the

directrix. The parabola is symmetric with respect to its axis.c. The latus rectum is the chord drawn though the focus and parallel to the directrix

(and therefore perpendicular to the axis) of the parabola.

d. In the parabola y2=4 ax , the length of latus rectum is 4 a , and the endpoints of the

latus rectum are (a , −2a ) and (a , 2a ) .

In the figure below, the vertex of the parabola is the origin, the focus is F (a , 0 ) , the directrix

is the line containing L1 L2 , the axis is the x -axis, the latus rectum is the line containing C1C2 .

Figure 10

Figure 11

E. ELLIPSE

1. Definition

An ellipse is the set of all points P on a plane such that the sum of the distances of P

from two fixed points F1 and F2 on the plane is constant. Each fixed point is called focus (plural: foci).


2. Equation of an ellipse

a. If the center is at the origin, the vertices are at (±a , 0 ) , the foci at (±c , 0 ) , the

endpoints of the minor axis are at (0 , ±b ) and b2=a2−c2 , then the equation is

x2

a2+ y

2

b2=1

.

Figure 12

b. If the center is at the origin, the vertices are at (0 , ±a ) , the foci are at (0 , ±c ) , the

endpoints of the minor axis are at (±b , 0 ) and b2=a2−c2 , then the equation is

x2

b2+ y

2

a2=1

.

Figure 13

c. If the center is at (h , k ) , the distance between the vertices is 2a , the principal axis is

horizontal and b2=a2−c2 , then the equation is

( x−h )2

a2+ ( y−k )2

b2=1

.

d. If the center is at (h , k ) , the distance between the vertices is 2a , the principal axis is

vertical and b2=a2−c2 , then the equation is

( y−k )2

a2+ ( x−h )2

b2=1

.

3. Parts of an ellipse

For the terms described below, refer to the ellipse shown with center at

O , vertices at

V 1 (−a , 0 ) and V 2 (a , 0 ) , foci at F1 (−c , 0 ) and F2 (c , 0 ) , endpoints of the minor axis at


x

y

O

B2(0,b)

B1(0,-b)

F1(-c,0) F2(c,0)V1(-a,0) V2(a,0)

),( 2abc

),( 2abc

),( 2abc

),( 2abc

B1 (0 , −b ) and B2 (0 , b ) , endpoints of one latus rectum at G1(−c ,−b2a )

and

G2(−c , b2a ) and the other at

H1(c ,−b2a ) and

H2(c , b2a ).

Figure 14

a. The center of an ellipse is the midpoint of the segment joining the two foci. It is the intersection of the axes of the ellipse. In the figure above, point O is the center.

b. The principal axis of the ellipse is the line containing the foci and intersecting the ellipse at its vertices. The major axis is a segment of the principal axis whose

endpoints are the vertices of the ellipse. In the figure, V 1V 2 is the major axis and has length of 2a units.

c. The minor axis is the perpendicular bisector of the major axis and whose endpoints

are both on the ellipse. In the figure, B1B2 is the minor axis and has length 2b units.

d. The latus rectum is the chord through a focus and perpendicular to the major axis.

G1G2 and H1H2 are the latus rectum, each with a length of

2b2

a .

Figure 15

The graph of 9 x2+16 y2=144 .


x

y

O(-4,0) (4,0) (5,0)

)59(4,

(-5,0)

)59(4,-

)59(-4,

)59(-4,-

(0, -3)

(0, 3)

x

y

O(12,1)

(2,-4)

(-8,1)

(2,6)

(-6,4)

(2,1)

(8,5)

(8,3)

1925

22

yx 1

25)1(

100)2( 22

yx

Figure 16

4. Kinds of ellipses

a. Horizontal ellipse. An ellipse is horizontal if its principal axis is horizontal. The graphs above are all horizontal ellipses.

b. Vertical ellipse. An ellipse is vertical if its principal axis is vertical.

F. HYPERBOLA

1. Definition

A hyperbola is the set of points on a plane such that the difference of the distances of each point on the set from two fixed points on the plane is constant. Each of the fixed points is called focus.

Figure 17

2. Equation of a hyperbola

a. If the center is at the origin, the vertices are at (±a , 0 ) , the foci are at (±c , 0 ) , the

endpoints of the minor axis are at (0 , ±b ) and b2=c2−a2 , then the equation is

x2

a2− y2

b2=1

.


Figure 18

b. If the center is at the origin, the vertices are at (0 , ±a ) , the foci are at (0 , ±c ) , the

endpoints of the minor axis are at (±b , 0 ) and b2=c2−a2 , then the equation is

y2

a2− x2

b2=1

.

Figure 19

c. If the center is at (h , k ) the distance between the vertices is 2a , the principal axis is

horizontal and b2=c2−a2 , then the equation is

( x−h )2

a2− ( y−k )2

b2=1

.

d. If the center is at (h , k ) , the distance between the vertices is 2a , the principal axis is

vertical and b2=c2−a2 , then the equation is

( y−k )2

a2− (x−h )2

b2=1

.

3. Parts of a hyperbola

For terms described on the next page, refer to the hyperbola shown which has its center

at O , vertices at V 1 (−a , 0 ) and V 2 (a , 0 ) , foci at F1 (−c , 0 ) and F2 (c , 0 ) and endpoints

of one latus rectum at G1(−c ,−b2a )

and G2(−c , b2a )

and the other at H1(c ,−b2a )

and H2(c , b2a )

.


x

y

O

B2(0,b)

B1(0,-b)

F1(-c,0) F2(c,0)

V1(-a,0)

V2(a,0)

),( 2abc

),( 2abc

),( 2abc

),( 2abc

x

y

O

(0,b)

(0,-b)

(-a,0) (a,0)

xa

by x

a

by

P

Figure 20

a. The hyperbola consists of two separate parts called branches.

b. The two fixed points are called foci. In the figure, the foci are at (±c , 0 ) .c. The line containing the two foci is called the principal axis. In the figure, the

principal axis is the x -axis.d. The vertices of a hyperbola are the points of intersection of the hyperbola and the

principal axis. In the figure, the vertices are at (±a , 0 ) .e. The segment whose endpoints are the vertices is called the transverse axis. In the

figure V 1V 2 is the transverse axis.

f. The line segment with endpoints (0 , b ) and (0 , −b ) where b2=c2−a2 is called the

conjugate axis, and is a perpendicular bisector of the transverse axis.g. The intersection of the two axes is the center of the hyperbola.h. The chord through a focus and perpendicular to the transverse axis is called a latus

rectum. In the figure G1G2 is a latus rectum whose endpoints are G1(−c ,−b2a )

and G2(−c , b2a )

and has length of

2b2

a .

4. The asymptotes of a hyperbola

Shown in the figure below is a hyperbola with two lines as extended diagonals of the rectangle shown.

Figure 21


x

y

F’(-6,0) O(-3,0) (3,0) F(6,0)

(6,9)

(6,-9)

x

y

F’(0,-6)

O

(0,3)

F(0,6)

(0,-3)

(-9,6) (9,6)

03 xy

03 xy

1279

22

yx

1279

22

xy

These two diagonal lines are said to be the asymptotes of the curve, and are helpful in sketching the graph of a hyperbola. The equations of the asymptotes associated with x2

a2− y2

b2=1

are y=bax

and y=−b

ax

. Similarly, the equations of the asymptotes

associated with

y2

a2− x2

b2=1

are y=abx

and y=−a

bx

.

Figure 22

The graph of 9 x2−16 y2=144 .

Figure 23


PART II – ANALYZING TEST ITEMS

Choose the letter of the best answer.

For items no. 1-7, please refer to the following points: J (3 ,−4 ) , Y (−2 , 5 ) , and H (0 , 3 ) .

1. What is the slope of JY ?

A.−59 B.

95 C.

− 95 D.

59

Solution:m=

5− (−4 )(−2 )−3

=−95 . Thus, the answer is C.

2. What is the equation of J⃗Y ?

A. 9 x+5 y−47=0 B. 5 x+9 y−7=0 C. 9 x−5 y+7=0 D. 9 x+5 y−7=0

Solution: The slope of J⃗Y is − 95 . Use this slope and J (3 ,−4 ) to find the equation

of the line.

y− (−4 )=−95

(x−3 )

5 y+20=−9 x+279 x+5 y−7=0 . The answer is D.

3. What is the equation of the line parallel to J⃗Y and passing through H ?

A. 9 x+5 y−15=0 B. 5 x+9 y−15=0 C. 9 x−5 y−15=0 D. 9 x−5 y+15=0

Solution: The slope of J⃗Y is −95 . Note that parallel lines have equal slopes.

Use the slope-intercept form to solve for the equation of the line.

y−3=−95

(x−0 )

5 y−15=−9 x9 x+5 y−15=0 . The answer is A.

4. What is the equation of the line perpendicular to J⃗Y and whose y -intercept is −10?

A. 9 x−5 y−20=0 B. C. D. 5 x−9 y−5=0

Solution: The slope of J⃗Y is − 95 . Therefore, the slope of the line perpendicular to

J⃗Y is

59 . Thus,

y= 59x−10

9 y=5 x=10−5 x+9 y+10=0 or 5 x−9 y−10=0 . The answer or B.


5. What is the distance of H from J⃗Y ?

A.

4 √7853 B.

√10653 C.

4 √10653 D.

2√10653

Solution: The equation of J⃗Y is 9 x+5 y−7=0 . The distance of H from J⃗Y is

d=|(9 ) (0 )+(5 ) (3 )+(−7 )|

√92+52=4|10653

. The answer is C.

6. What are the coordinates of the midpoint of YH ?

A. (−1 , 4 ) B. (−1 ,−4 ) C. (1 , 4 ) D. (1 ,−4 )

7. What are the coordinates of P , which divides YH in the ratio 2 :5?

A.(−317 , 107 )

B.(107 , −317 )

C.(107 , 317 )

D.(−107 , 317 )

8. What is the equation of the line whose x -intercept is −10 and y -intercept is 8 ?

A. 4 x+5 y=0 B. 4 x−5 y+40=0 C. 5 x−4 y+40=0 D. 4 x+5 y−10=0

9. Which of the following is the equation of the perpendicular bisector of the line segment

joining the points X (2 , 4 ) and B (8 ,−2 ) ?A. x+ y−4=0 B. x−2 y−9=0 C. x− y−4=0 D. 6 x− y+4=0

10. Determine the center of the circle whose equation is 4 x2+4 y2−16 x−24 y+51=0 .

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

11. What is diameter of the circle x2+ y2−6 x+4 y=12?

A. 4 B. 5 C. 10 D. 16

12. What is the equation of the circle whose diameter is the segment with endpoints K (4 , 3 ) and M (20 ,−9 ) .A. ( x−6 )2+( y+3 )2=400 C. ( x+12 )2+( y−3 )2=100

B. ( x−3 )2+( y+12 )2=100 D. ( x−12 )2+ ( y+3 )2=100

13. Line y=2x+3 is tangent to a circle with center (2 ,−3 ) . Which of the following is the length of the radius of the circle.

A. 2√5 B. √5 C. 2 .5 D. 5√2

14. Determine an equation for the parabola that opens upwards with focus 4 units from the vertex.

A. x2=−4 y B. y

2=4 x C. x2=16 y D. x

2=−16 y

15. Determine the focus of the parabola y2=32x−64 .


A.(2 , 18 ) B. (10 , 0 ) C. (0 , 10 ) D. (−2 , 32 )

16. Determine the equation of the directrix of the parabola .

A. x=−12 B. y=8 C. y=−20 D.

17. Determine the equation of the directrix of the parabola x2 + 6x + 20y - 71 = 0.

A. y = -6 B. y = 9 C. x = 9 D. x = -6

18. Determine the length of the latus rectum of the parabola x2 + 6x + 20y - 71 = 0.

A. 6 B. 4 C. 5 D. 20

19. Determine the equation of the ellipse with center at the origin, one focus at (-12, 0) and length of the semi-major axis of 13.

A. B. C.

x2

144+ y2

169=1

D.

x2

169+ y 2

144=1

20. Determine the center of the ellipse whose equation is

.

A. (-4, -5) B. (4, 5) C. (-4, 3) D. (4, -3)

21. Which of the following is a focus of the ellipse whose equation is

.

A. (1, -3) B. (4, 1) C. (1, 4) D. (-3, 1)

22. Determine the length of the major axis of the ellipse

.

A. 5 B. 10 C. 8 D. 6

23. Determine the equation of the hyperbola having its foci at (0, 13) and length of the transverse axis of 10.

A. B. C. D.


PART III – ENHANCING TEST TAKING SKILLS

24. Which of the following is a vertex of the hyperbola having its foci at (0, 13) and length of the transverse axis of 10?

A. (0, -5) B. (10, 0) C. (0, 10) D. (12, 0)

25. Which of the following is a focus of the hyperbola ?

A. (-5, -12) B. (3, -20) C. (13, 4) D. (-7, 4)

Choose the letter of the best answer.

1. What is the distance between the points T (4 , −9 ) and K (8 , 4 ) ?A. 13 B. √185 C. √119 D. 24

2. What is the slope of T⃗K if T ( 4 , −9 ) and K (8 , 4 ) ?

A.

−413 B.

−134 C.

413 D.

134

3. Determine an equation of the line passing through the point (0 , 2 ) with slope 8.

A. y=8x+2 B. y=2x+6 C. y=8x−6 D. y=8x+6

4. Determine an equation of the line passing though the points (2 , 11) and (15 , 2 ) .A. 4 y−9 x−145=0 C. 13 y+9 x−161=0B. 13 y−9 x+161=0 D. 9 y+13 x−150=0

5. Which is of the following is the equation of the line passing through the point (16 , 15 ) and

perpendicular to y=−4?

A. x=16 B. y=4 C. x=−16 D. x=4

6. What is the equation of the directrix of the parabola y2+2 y+12 x+25=0?

A. y=−3 B. y=3 C. x=−1 D. x=1

7. Determine the focus of the parabola y+12 x−2x2=16 .

A.(−3 , −158 )

B.(3 , −158 )

C.(158 , 3) D. (3 , −9 )

8. Determine the equation of the parabola with vertex at (1 , 0 ) and directrix x=−5 .


A. x2=24 ( y+1 ) B. y2=−24 ( x+1 ) C. x

2=24 ( y−1 ) D. y2=24 ( x−1 )

9. Determine the equation of the parabola with focus (3 , 6 ) and vertex (3 , 2 ) .A. x

2−6 x−16 y−41=0 C. x2−16x−6 y−41=0

B. y2−6 y−16 x−41=0 D. x2−6 x+16 y+41=0

10. Determine the latus rectum of the parabola x−7 y2=0

A. 7 B. √7 C.

17 D.

√77

11. Determine the coordinates of the foci of the ellipse x2+16 y2=1 .

A.(0 , ±√15

4 )B.

(±√154, 0)

C.(0 , ± 73 ) D.

(± 73 , 0)

12. Determine the eccentricity of the ellipse

15x2+ 1125

y2=5

A. 5 B.

15 C.

5√612 D.

√245

13. Determine the equation of the ellipse with foci (±5 , 0 ) and a major axis of length16 .

A.

x2

64+ y

2

39=1

B.

x2

39+ y

2

64=1

C.

x2

39+ y2

256=1

D.

x2

36+ y

2

64=1

14. Determine the equation of the ellipse with center (2 , 2 ) , focus , and vertex

A.

( x−2 )2

5+

( y−2 )2

9=1

C.

( x−2 )2

9+

( y−2 )2

5=1

B.

( x−9 )2

2+

( y−5 )2

2=1

D.

( x−5 )2

2+

( y−9 )2

2=1

15. Determine the equation of the ellipse with foci (±2 , 0 ) and vertices (±5 , 0 ) .

A.

x2

21+ y

2

25=1

B.

x2

25+ y

2

21=1

C.

x2

5+ y

2

7=1

D.

x2

7+ y

2

5=1

16. Find the foci of the hyperbola 16 x2−25 y2=400 .

A.(0 , ±√13 ) B. (±√13 , 0 ) C. (0 , ±√41 ) D. (±√41 , 0 )

17. What is the length of the conjugate axis of the hyperbola

?

A. 9 B. 8 C. 4 D. 3

18. Find the equation of the hyperbola with foci (±10 , 0 ) and vertices (±8 , 0 ) .


A.

x2

64− y2

36=1

B.

x2

36− y2

64=1

C.

x2

8− y2

6=1

D.

x2

6− y2

8=1

19. Determine the equation of the hyperbola with vertices (0 , ±4 ) that passes through (3 , 5 ) .

A.

x2

4− y2

4=1

B.

y2

4− x2

4=1

C.

x2

16− y2

16=1

D.

y2

16− x2

16=1

20. Determine the equation of the hyperbola with foci at (10, 10) and length of its transverse axis of 12.

A. C.

B. D.

21. Given the center of a circle at (2 ,−4 ) and one of the points on the circle is (5 , −8 ) , what is the length of the diameter of the circle?A. 4 units B. 8 units C. 10 units D. 12 units

22. What is the equation of a circle with center at (4 ,−6 ) and radius of 5?

A. x2+ y2−8 x+12 y+27=0 C. x

2+ y2−12x+8 y+30=0

B. x2+ y2−8 x−12 y−27=0 D. 2 x

2+2 y2+8 x−12 y+27=0

23. What is the equation of the circle with center (3 , 5 ) that passes through the point (−4 , 10 ) ?A. 3 x

2+3 y2−15x−6 y−30=0 C. x2+ y2−10 x−6 y−30=0

B. x2+ y2−5 x−3 y−15=0 D. x

2+ y2−6 x−10 y−40=0

24. What is the radius of the circle 2 x2+2 y2+24 x−81=0?

A.

√115 B.

√343 C.

3√342 D.

4 √67

25. What is the center of the circle x2+ y2−8 x+2 y+8=0 ?

A. (−1 , 4 ) B. (4 ,−1 ) C. (−4 , 1 ) D. (1 ,−4 )

Answer Key


Part II – Analyzing Test Items

1. C 6. A 11. C 16. D 21. A2. D 7. D 12. D 17. B 22. B3. A 8. B 13. A 18. D 23. C4. B 9. C 14. C 19. B 24. A5. C 10. A 15. B 20. D 25. A

Part III - Enhancing Test taking Skills

1. B 6. C 11. B 16. D 21. C2. D 7. B 12. D 17. B 22. A3. A 8. D 13. A 18. A 23. D4. C 9. D 14. A 19. D 24. C5. A 10. C 15. B 20. C 25. B


Documents

Analytic Geometry _2012