ANALYTIC GEOMETRY
Distance Between Two Points xL
xR
yH
Horizontal distance = x R – xL yL
Vertical distance = y H – yL y
P 2(x2,y2 )
P 1(x1,y1)
d = x
( x2 − x1 ) 2 + ( y 2 − y1 ) 2
The Straight Line Slope of a Line: y
P2(x2,y2)
y 2 – y1
α P1(x1,y1)
x 2 – x1
x
m = tan α
⇒
m=
y 2
− y1
x 2
− x1
The slope, m is positive if the t he line is inclined upwards to the right but is negative if the line is inclined downwards downwards to the right. For horizontal lines, m = 0 and for vertical lines, it is undefined, i.e m = ∞ .
For parallel lines, m1 = m2 ; for perpendicular perpendicular lines, m1
=−
1 . m2
Midpoint of a Line Segment: y
P 2(x2,y2)
Let ( x, y ) be the coordinates of the t he midpoint of the line segment P 1P2.
M ( x , y )
x =
P1(x1,y1)
x1 + x 2 2
x
y L2
φ
Angle Between Two Lines: L1
x
y=
y1 + y 2 2
Let be the angle measured counterclockwise from line L1 to line L2. m − m1 tan φ = 2 1 + m1 m 2 where m1 – slope of initial side, i.e. line L 1 m2 – slope of terminal side, i.e. line L 2
Division of a Line Segment : y
Let P(x,y) be a point located on the line joining P1(x1,y1) and P2(x2,y2) such that P1P is a given fraction k of P1P2, i.e.
P2(x 2,y 2)
P(x,y)
P1P = k P1P2 P1(x1,y1)
Then
x
x = x1 + k (x2 – x1) y = y1 + k (y2 – y1)
Note that k is positive if the directed line segments P1P and P1P2 are similarly directed and negative if P1P and P1P2 are oppositely directed. Equations of Straight Lines: General equation:
Ax + By + C = 0
where A, B & C are real numbers
Point-Slope Form:
y – y1 = m (x – x1) a given
where (x1, y1) are the coordinates of
point Two-Point Form: y – y1 = Slope-Intercept Form:
y2
− y1
x 2
− x1
(x – x1) where (x1, y1) & (x2, y2) are given points
y = mx + b
where b – y-intercept of the line Intercept Form:
x
+
y
y
b p
=1
a b where a – x-intercept of the line
β a
x
x cos β + y sin β = p Normal Form: where p – distance of the line from the origin
To reduce the equation Ax + By + C = 0 into the normal form, divide through by ± A 2 + B 2 (using the sign in front of B), then transpose the constant term to the right, i.e.
Ax ±
A 2
By
+
+ B
2
±
A 2
− C
=
+ B
2
A 2
±
+ B
2
= p
Note that p can be positive or negative. If p is positive, it means that the line Ax + By + C = 0 is above (or to the right of) the origin while if p is negative, it means that the line is below (or to the left of) the origin. Distance Between Two Parallel Lines: y Ax+By+C2=0
d
d = x
C 2 A 2
−
C 1 B 2
+
Ax+By+C 1=0
Distance from a Point to a Line:
y
The perpendicular distance from a point P 1(x1, y1) to the line Ax + By + C = 0 is given by
Ax + By + C = 0
d
d =
Ax1
+ By1 + C
A 2
±
+ B
P1(x1,y1)
2
x
As in the normal form of a line, use the sign in front of B in the denominator. Hence, d could come out either positive or negative. If d is positive, it means that the point is above the line while if it is negative, it means that the point is below the line. Area of a Triangle: y
For a triangle with given vertices,
P3(x 3,y 3)
P1(x 1,y 1)
A = P2(x 2,y 2)
1 2
x1
y1
1
x 2
y 2
1
x3
y 3
1
x
Note that for the computed area to be positive, the vertices should be numbered in the counterclockwise order
The area formula is sometimes written in the more convenient form shown below. A =
1 2
x1
x 2
x 3
x1
y1
y 2
y 3
y1
1 =
2
[ x y 1
2
x 2 y 3
+
+
x3 y1
x1 y 3
−
x3 y 2
−
x 2 y1 ]
−
For a polygon of n sides whose vertices are known, A =
1
x1
x 2
x 3
.
.
.
.
x n
x1
2
y1
y 2
y 3
.
.
.
.
y n
y1
Examples: 1. Find the area of the pentagon having vertices at (3. 0), (2, 3), (–1, 2), (–2, –1) and (0, –2). Solution: y P3(-1,2)
Numbering the vertices in the counterclockwise order, the area is
P 2(2,3)
P 1(3,0) P4(-2,-1)
x
A
P 5(0,-2)
=
1 3
2
−
1
2 0
3
2
−
2
1
−
0 −
2
3 0
= 1 [9 + 4 + 1 + 4 + 0 − ( −6 − 0 − 4 − 3 − 0) ] 2 =
31 2
square units
2. What is the equation of a line that passes through (4, 0) and is parallel to the line x – y – 2 = 0? Solution: The equation of the required line is x – y + k = 0. To find k, substitute the coordinates of the given point. Hence, 4 – 0 + k = 0. Hence, k = –4. ∴x – y – 4 = 0 3. Find the equation of the line through (3, 1) that is perpendicular to the line x + 5y + 5 = 0. Solution: The equation of the required line can be obtained by interchanging the coefficients of x and y from the given equation and changing the sign of one of them, i.e. 5x – y + k = 0 ⇒ 5(3) – 1 + k = 0 ⇒ k = –14 ∴5x – y –14 = 0 THE CONIC SECTIONS A conic section is the locus of a point that moves such that its distance from a fixed point (called focus) is in constant ratio to its distance from a fixed straight line (called directrix ). This constant ratio is called eccentricity . The circle, parabola, ellipse and hyperbola compose the conic sections (or simply conics) since each of them can be formed by a plane which is made to intersect a cone.
The Circle: A circle is the locus of a point in a plane that moves so that it is always equidistant from a fixed point (called the center ). The fixed distance is called the radius. A circle is produced when the cutting plane is parallel to the base of t he cone.
Equations of a Circle: General equation: Ax2 + Ay2 + Dx + Ey + F = 0 or x2 + y2 + Dx + Ey + F = 0 Standard form:
(x – h)2 + (y – k)2 = a2
where (h, k) – coordinates of the center and a – radius of the circle
The Parabola: A parabola is the locus of a point in a plane that moves such that its distance from a fixed point (called the focus) equals its distance from a fixed line (called the directrix ). It is formed when the cutting plane is parallel to one of the elements of the cone. Since the parabola is a symmetrical curve, the line of symmetry is called the axis of the parabola. Equations of a Parabola: General equations: Vertical axis: Ax2 + Dx + Ey + F = 0 or x2 + Dx + Ey + F = 0 Horizontal axis: Cy2 + Dx + Ey + F = 0 or y2 + Dx + Ey + F = 0 Standard Forms: (x – h)2 = ± 4a (y – k)
Vertical axis:
where (h, k) – coordinates of the vertex a – undirected distance from vertex to focus, i.e. a > 0.
If the right side of the equation is positive, the parabola opens upward; if the right side of the equation is negative, the parabola opens downward. Horizontal axis: (y – k)2 = ± 4a (x – h) If the right side of the equation is positive, the parabola opens rightward; if the right side of the equation is negative, the parabola opens leftward. y
Latus rectum, LR = 4a Directrix 2a
a
V
a
F
Axis x
2a D
P
e
=
PF PD
For a parabola, e = 1.
The Ellipse: An ellipse is the locus of a point in a plane that moves such that the sum of its distances from two fixed points (called the foci ) is a constant and is equal to the length of the major axis (2a). It can also be defined as the locus of a point that moves such that the ratio of its distance from a fixed point (or focus) to its distance from a fixed line (or directrix ) is a constant and is less than one. The ellipse is the conic section formed if the cutting plane is making an angle (other than 90o) with the axis of the cone. Equations of an Ellipse: General equation: Ax2 + Cy2 + Dx + Ey + F = 0 Standard Forms: Center at (h, k), Major axis horizontal
Major axis
y
a
( x − h)
a
2
+
2
( y − k ) b
2
=
2
x
1 y
Center at (h, k), Major axis vertical
( x − h)
b
2
+
2
( y − k ) a
Major axis
2
=
2
where a – semimajor axis b – semiminor axis y
x
1 ⇒
⇒
Major axis = 2a Minor axis = 2b Directrix
Latus rectum , LR = 2b2/a
b V2
F2
C
F1
a
a
x
b D2
P
V1
c d
Note that a > b.
Referring to the figure above , c – distance from center to focus such that c2 = a2 – b2 a d – distance from center to directrix such that d = e e – eccentricity of the ellipse
e=
PF 2 PD2
=
c a
<1
The Hyperbola: The hyperbola is the locus of a point in a plane that moves such that the difference of its distances from two fixed points (called foci ) is a positive constant and is equal to the length of the transverse axis (2a). It can also be defined to be the locus of a point that moves such that the ratio of its distance from a fixed point (or focus) to its distance from a fixed line (or directrix ) is a constant and is greater than one. The hyperbola is the conic section formed if the cutting plane is parallel (but not coincident) to the axis of the cone. Equations of the Hyperbola: General equation : Ax2 + Cy2 + Dx + Ey + F = 0 where A & C have unlike signs
Standard Forms: Transverse axis – the line segment joining the two vertices of the hyperbola which has a total length of 2a. Center at (h, k), Transverse axis - horizontal: y
Transverse axis
x
( x − h )
2
−
a2
( y − k )
2
=1
b2
Center at (h, k), Transverse axis - vertical:
y Transverse axis
( y − k )
2
−
a2
( x − h )
2
=1
b2
O
x
where a – semitransverse axis ⇒ Transverse axis = 2a b – semiconjugate axis
⇒ Conjugate axis = 2b
c – distance from center to focus such that c2 = a2 + b2 a d – distance from center to directrix such that d = e e – eccentricity of the hyperbola
e=
PF 2
=
PD2
c a
>1
y Directrix
d
Latus rectum, LR = 2b2/a
b F2 V2
V1 F1 a
P
C b
a
x
D2 c
Asymptote
Degenerate conic (point circle, one line, two intersecting lines) – the conic formed if the cutting plane passes through the vertex of the cone. Examples: 1. Find the value of k for which the equation x2 + y2 + 4x – 2y – k = 0 represents a point circle.
2.
a. 5 b. 6 c. –6 d. –5 Solution: Reducing the given equation into the standard form (x2 + 4x + 4) + (y2 – 2y + 1) = k + 4 + 1 (x + 2)2 + (y – 1)2 = k + 5 For a point circle, the radius, a = 0 ⇒ a2 = 0 ⇒ k = –5 ∴k + 5 = 0 2 2 The equation x – y – 4x – 6y – 5 = 0 represents a. a circle b. an ellipse c. a hyperbola d. two straight lines Solution: Reducing the given equation into the standard form (x2 – 4x + 4) – (y2 – 6y + 9) = 5 + 4 – 9 (x – 2)2 – (y – 3)2 = 0 [(x – 2) + (y – 3)][(x – 2) – (y – 3)] = 0 (x + y – 5)(x – y + 1) = 0 x+y–5=0 x–y+1=0 Therefore, the given equation represents two straight lines.
General Equation Of Second Degree: This is also the general equation of a conic section: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 When B ≠ 0, the principal axes of the conic are inclined by an angle θ expressed as B tan 2θ = A − C In other words, the principal axes are not parallel to the coordinate axes. The curve can be obtained from the equation given by determining the value of the discriminant B2 – 4AC. If B2 – 4AC < 0, the conic is an ellipse (e < 1). If B2 – 4AC = 0, the conic is a parabola (e = 1). If B2 – 4AC > 0, the conic is a hyperbola (e > 1). Examples: 1. Using the discriminant, identify the curve represented by the equation 3x2 + 6xy + 3y2 – 4x + 7y = 12 Solution: For the given equation, A = 3, B = 6 and C = 3. B2 – 4AC = 62 – 4(3)(3) = 0. Hence the equation represents a parabola. 2. Find the angle of rotation that will eliminate the xy term in the equation 3x2 + 6xy + 3y2 – 4x + 7y = 12 Solution: B 6 6 = = = ∞ ⇒ 2θ = tan −1 ∞ = 90 ° ⇒ θ = 45o tan 2θ = A − C 3 − 3 0 When B = 0, the principal axes of the conic are parallel to the coordinate axes, i.e. the x- and the y-axes.
Polar Coordinate System: In this system, the location of a point in a plane is expressed by the ordered pair (r, θ) where r (called the radius vector ) is the distance of the point from the origin (or the pole) and θ θ) (called the polar angle) is the angle that the radius vector makes with the Ox(r, axis (also called the polar axis). r
θ Polar axis
x
Sign Convention: • θ is positive (+) when measured counterclockwise • θ is negative (–) when measured clockwise • r is positive (+) when laid off at the terminal side of θ • r is negative (–) when laid off at the extension of the terminal side of θ in the opposite quadrant Relationship Between Rectangular and Polar Coordinates: An equation in Cartesian (or rectangular) coordinates can be t ransformed into one in polar coordinates and vice-versa. The following equations can be obtained from the diagram shown. Conversion from rectangular to polar: (r, θ) (x, y)
y
r
x = r cos θ
y
y = r sin θ
Conversion from polar to rectangular:
θ x
x
r = x 2
+
tan θ =
y 2
x
cos θ =
x 2
+
sin θ =
y 2
y x y x 2
+ y
Examples: 1.
What is the polar equation of a vertical line 5 units to the right of the origin? Solution: y
By inspection, the equation of the vertical line in rectangular coordinates is
x=5
x=5 Hence, in polar form, the equation is
x
r cos θ = 5
2. Express the polar equation r = 4a sin θ in rectangular form. Solution: r = 4a sin θ
⇒
x 2
+ y
Hence, the required equation is
2
=
y
4a
x
2
+ y
2
⇒
x 2
+
y 2
=
4ay
2
x 2
+
y 2
−4
ay
=0
Review Exercises in Analytic Geometry 1. The rectangular coordinate system for a plane is divided into four parts which are known as a. coordinates b. octants c. quadrants d. axes 2. When two lines are perpendicular, the slope of one is a. equal to the other b. equal to the negative of the other c. equal to the reciprocal of the other d. equal to the negative reciprocal of the other 3. If the product of the slopes of any two straight lines is negative 1, then the lines are a. parallel b. askew c. perpendicular d. non-intersecting 4. It represents the distance of a point from the y-axis. a. ordinate b. coordinate c. abscissa
d. polar distance
5. A line perpendicular to the x-axis has a slope of a. zero b. unity c. infinity
d. none of these
6.
In finding the distance between two points P 1(x1, y) and P 2(x2, y2), the most direct procedure is to use a. the Law of Cosines b. the slope of the line c. translation of axes d. the Pythagorean Theorem
6. In the triangle ABC having coordinates of A(-2, 5), B(6, 1) and C(-2, -3), find the length of the median from the vertex B to the line AC. a. 8 b. 6 c. 7 d. 5 8. Find the coordinates of the centroid of the triangle ABC with coordinates of A(0. -3), B(3, 0) and C(0, -6). a. (2, -2) b. (1, -3) c. (1, -2) d. (2, -3) 9. If (x, 4) is equidistant from (5, -2) and (3, 4), find x. a. 10 b. 12 c. 13
d. 15
10. Find x if the distance between points (x, 4) and (3, 4) is equal to 10. a. 12, -6 b. 11, -4 c. 14, -7
d. 13, -7
11. The line segment connecting (x, 6) and (9, y) is bisected by the point (7, 3). Find the value of x and y. a. 4, 0 b. 5, 2 c. 4, 1 d. 5, 0 12. Find the equation of the perpendicular bisector of the line joining (4, 0) and (-6, 3). a. 20x + 3y + 28 = 0 b. 10x + 20y + 24 = 0 c. 20x – 6y + 29 = 0 d. 10x + 4y + 24 = 0 13. Find the smallest angle between the lines 2x + y – 8 = 0 and x + 3y + 4 = 0. a. 60o b. 45o c. 30o d. 40o 14. Find the area of the polygon whose vertices are at (2, -6), (4, 0), (2, 4), (-3, 2) and (-3, -3). a. 47.5 b. 55.3 c. 45.2 d. 57.4 15. The point B(-4, 1) is three fifths of the distance from one end A(2, -2) of a line segment to the other end C(x, y). Find end point C(x, y).
a. (-8, 3)
b. (-2, 0)
c. (-5, 2)
d. (-6, 3)
16. Find the equation of a line with slope 3 and y-intercept -2. a. y = 2x – 3 b. y = 3x – 2 c. y = -3x + 2
d. y = 2x + 3
17. The distance from the point (2, 1) to the line 4x – 3y + 5 = 0. a. 1 b. 4 c. 3
d. 2
18. A line passes through (4, 2) and has a slope of 2. Find the equation of the line. a. 2x – 3y + 10 = 0 b. 3x + 2y – 10 = 0 c. 2x – y – 6 = 0 d. 3x – 2y + 8 = 0 19. Find the equation of a line having an x-intercept of 2 and a y-intercept of 4. a. 2x + y = 4 b. x + 2y = 4 c. x – 2y – 4 = 0 d. 2x – y + 4 = 0 20. Find the equation of a line through (4, 6) perpendicular to the x-axis. a. y – 6 = 0 b. x – 4 = 0 c. x + y = 0
d. y – 4 = 0
21. Find the equation of the line through (2, 4) and (4, 6). a. x + y + 4 = 0 b. x – 2y + 2 = 0 c. x – y + 2 = 0
d. 2x – y + 4 = 0
22. A line passes through (2, 4) and perpendicular to the line 3x + 4y – 4 = 0. Find the equation of the line. a. 3x + 4y + 20 = 0 b. 2x + 3y – 18 = 0 c. 3x + 2y – 18 = 0 d. 4x – 3y + 4 = 0 23. Find the equation of a straight line with a slope of 2 and x-intercept of 1. a. 2xs + y + 2 = 0 b. 2x – y – 2 = 0 c. x – 2y + 2 = 0 d. 2x – y + 2 = 0 24. Find the equation of a line passing through (1, 4) and parallel to the line 3x – 4y = 6. a. 3x + 4y – 13 = 0 b. 3x – 4y + 13 = 0 c. 4x – 3y + 13 = 0 d. 4x + 3y – 13 = 0 25. Find the distance between the given lines 4x – 3y = 12 and 4x – 3y = -8. a. 4 b. 10 c. 8 d. 3 26. What is the slope of the line 3x + 2y + 1 = 0? a. 2/3 b. -2/3
c. -3/2
d. 3/2
27. How far is the intersection of the lines 4x – 5y = 26 and 3x + 7y +2 = 0 from the origin? a. 3 5 b. c. 4 5 d. 2 5 5 28. Find the value of k so that the lines kx + 3y – 6 = 0 and 9x – 6y + 1 = 0 will be perpendicular. a. 2 b. 3 c. 4 d. 5 29. A line passes through (-2, 5) and (k, 1) and has x-intercept of 3. Find k. a. 2 b. 3 c. 1 d. 4 30. Find the x-intercept of a line which is perpendicular to a line 3x + 4y + 8 = 0 and which passes through point (2, 1). a. 2/3 b. 5/4 c. ¾ d. 1/3 31. The general second degree equation has the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. If B2 – 4AC = 0, the equation describes a. a circle b. an ellipse c. a hyperbola d. a parabola
32. Find the equation of the circle circumscribing a triangle whose vertices are at (0, 0), (0, 5) and (3, 3). a. x2 + y2 – x – 5y = 0 b. x2 + y2 – x – 4y = 0 c. x2 + y2 – 2x – 3y = 0 d. x2 + y2 – 2x – 6y = 0 33. All circles having the same center but with unequal radii are called a. tangent circles b. unequal circles c. eccentric circles
d. concentric circles
34. Find the equation of the circle tangent to the y-axis and having the center at (5, 3). a. (x – 3)2 + (y - 3)2 = 25 b. (x – 5) 2 + (y - 3)2 = 25 c. (x – 4)2 + (y - 5)2 = 20 d. (x – 5) 2 + (y - 5)2 = 25 35. The general second degree equation has the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0. If
B2 – 4AC = 0 describes an ellipse if a. B2 – 4AC = 0 b. B2 – 4AC = 1
c. B2 – 4AC > 0
d. B2 – 4AC < 0
36. Find the equation of the circle whose center is at (1, -3) and which passes through (-3, 5). a. (x + 1) 2 + (y +3)2 = 100 b. (x – 2) 2 + (y - 3)2 = 80 c. (x – 1) 2 + (y +3)2 = 80 d. (x – 2) 2 + (y + 4) 2 = 100 37. Find the equation of the circle tangent to the line 3x + 4y = 15 and having its center at (-3, -4). a. (x + 3) + (y + 4) = 64 b. (x + 3) 2 + (y + 4) 2 = 64 c. (x + 5)2 + (y - 4) 2 = 64 d. (x – 3) 2 + (y - 4)2 = 64 38. How far is the center of the circle x 2 + y2 – 8x + 1 = 0 from the x-axis?
a. 2
b. 0
c. 1
d. 3
39. Compute the ratio of the area of the circle x 2 + y2 – 10x – 24y + 25 = 0 and the circle
x2 + y2 – 10x + 4y – 7 = 0. a. 5 b. 3
c. 6
d. 4
40. Find the distance between the centers of the two circles x 2 + y2 – 4x – 4y + 4 = 0 and
x2 + y2 – 4x + 8y + 4 = 0. a. 4 b. 6
c. 8
d. 10
41. Find the equation of a circle circumscribing the triangle formed by the lines y = 0, y = x and 2x + 3y = 10. a. x2 + y2 – 5x + y = 0 b. x2 + y2 – 4x + y = 0 c. x2 + y2 – 3x + y = 0 d. x2 + y2 – 6x + y = 0 42. Find the area of the curve 5x 2 + 5y2 + 10x – 5y + 3 = 0.
a. 4.02 sq. units
b. 2.04 sq. units
c. 3.22 sq. units
d. 1.96 sq. units
43. The shortest distance from A(3, 8) to the circle x 2 + y2 + 4x – 6y = 12 is equal to
a. 4.1
b. 1.82
c. 2.07
d. 3.21
44. Compute the length of the latus rectum of the parabola y 2 + 8x – 6y + 25 = 0.
a. 6
b. 8
c. 10
d. 12
c. ellipse
d. parabola
45. 4x2 – y2 = 16 is the equation of a
a. hyperbola
b. circle
46. The location of the focus of the parabola x 2 – 6x – 12y – 51 = 0 is at
a. (2, -3)
b. (3, -5)
c. (2, -5)
d. (3, -2)
47. The equation 9x2 + 16y2 + 54x – 64y = -1 describes
a. a hyperbola
b. a sphere
c. a circle
d. an ellipse
48. Find the vertex of a parabola having its focus at (7, -4) and its directrix y = 2. a. (-4, 2) b. (3, -5) c. (7, -1) d. (6, -2) 49. The sum of the distances from the two foci to any point in a/an ________ is a constant. a. parabola b. any conic c. hyperbola d. ellipse 50. Find the directrix of the parabola x 2 + 2x + 12y + 37 = 0.
a. y = 1
b. y = 0
c. x = 0
d. x = 1
51. What kind of conic has an equation of Ax 2 + Cy2 + Dx + Ey + F = 0.
a. circle
b. parabola
c. ellipse
d. hyperbola
52. Find the equation of a parabola with axis vertical and passing through (0, 0), (1, 0) and (5, -20). a. x2 + x + y 2 = 0 b. x2 – x + y = 0 c. x – x 2 + y2 = 0 d. x – x2 + y = 0 53. The equation x2 + Bx + y 2 + Cy + D = 0 is
a. a hyperbola
b. an ellipse
c. a parabola
d. a circle
54. An arch 18 m high has the form of a parabola with vertical axis. The length of a horizontal beam placed across the arch 8 m from the top is 64 m. Find the width of the arch at the bottom. a. 96 m b. 100 m c. 87 m d. 70 m 55. If eccentricity is less than one, then the curve is a. parabola b. ellipse c. hyperbola
d. circle
56. What is the area enclosed by the curve 9x 2 + 25y2 – 225 = 0?
a. 188.496 sq. units
b. 47.124 sq. units
57. The semiconjugate axis of the hyperbola
a. 3
b. -3
x 2 9
c. 150.796 sq. units −
y2
=1
4 c. 2
d. 73.398 sq. units
is d. -2
58. The orbit of the earth is an ellipse with the sun at a focus, the semimajor axis is 93 million
miles, the eccentricity is 1/60. Find the greatest and least distance of the earth from the sun in miles. a. 94.55, 91.45 b. 90.44, 95.36 c. 92.36, 86.23 d. 96.32, 100.31 2 2 59. Compute the eccentricity of the curve 9x + 4y – 24y – 72x + 144 = 0. a. 0.62 b. 0.75 c. 0.58 d. 0.84 2 2 x y 60. The length of the latus rectum for the ellipse + = 1 is equal to 64 16 a. 3 b. 4 c. 5 d. 6 61. Compute the length of the latus rectum of the hyperbola 9x 2 – 4y2 – 90x + 189 = 0
a. 6
b. 7
c. 8
d. 9
62. Find the equation of the asymptotes for the hyperbola (y – 5) 2 – (x + 5) 2 = 36.
a. y = ±x
b. y – 5 = ± (x + 5)
c. y – 4 = ± (x – 4)
d. y – 5 = ± (x – 5)
63. __________ is the locus of a point that moves in a plane so that the difference of its distances from two fixed points on the plane is constant. a. Hyperbola b. Ellipse c. Circle d. Parabola 64. In an ellipse, a chord which contains a focus and is perpendicular to the major axis is the a. focal width b. conjugate axis c. minor axis d. latus rectum 65. The line passing through the focus and perpendicular to the directrix of a parabola is called a. latus rectum b. axis of the parabola c. tangent line d. secant line 66. The axis of the hyperbola through its foci is known as a. conjugate axis b. transverse axis c. major axis
d. minor axis
67. The axis of the hyperbola, which is parallel to its directrices, is known as a. conjugate axis b. transverse axis c. major axis d. minor axis
ANSWER KEY: c d c c c d a 8. b 9. c 10. d 1. 2. 3. 4. 5. 6. 7.
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d c b a a b d c a b
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
c d b b a c d a a b
31. 32. 33. 34. 35. 36. 37. 38. 39. 40.
d a d b d c b c d b
41. 42. 43. 44. 45. 46. 47. 48. 49. 50.
a b c b a d d c d b
51. 52. 53. 54. 55. 56. 57. 58. 59. 60.
c b d a b b c a b b
61. 62. 63. 64. 65. 66. 67.
d b a d b b a
