Expert: Scott A Wilson - 4/4/2010

Question
1. Find and equation of the ellipse having the given points as foci and the given sum of the foci radii.
(-9,0); (9,0); 30

a)(x^2/15)+ (y^2/12)=1
b)(x^2/225)+(y^2/144)=1
c)(x^2/15)-(y^2/12)=1
d)(x^2/225)-(y^2/144)=1

2. Choose the correct coordinates of the foci:
x^2+4y^2-16=0

a)(12,0)and (-12,0)
b)(-2(3^1/2),0) and (2(3^1/2),0)
c)(-12^1/2,0) and (12^1/2,0)
d)(-4(3^1/2),0) and (4(3^1/2),0)


3. Find the equation of the hyperbola described:
Foci (0,-8) and (0,8); difference of focal radii 10

a)(y^2/25)-(x^2/39)=1
b)(y^2/25)+ (x^2/39)=1
c)(y^2/5)-(x^2/39)=1
d)(y^2/5)+(x^2/39)=1

4. Find an equation of the hyperbola described:
Asymptotes y=x and y=-x; foci (-4,0) and (4,0)

a) (x^2/8)-y^2/8)=1
b) (x^2/4)+(y^2/4)=1
c)x^2-y^2=1
d)x^2+y^2=1


5. Use the information to write an equation of the porabola:
Focus:(-8,-5), Vertex: (2,-5)

a)(y+5)^2=-40(x-2)
b) (x+5)^2=4(y-2)
c) (y-5)^2=-4(x-2)
d) (x-5)^2=40(y-2)

6. Choose the correct foci:
5x^2+9y^2=45

a)(+/-2,0)
b)(+/-4,0)
c) (0,+/-2)
d) (0,+/-4)


7. Choose the correct foci:
9x^2+4y^2=9

a) (0,+/-2.5)
b) (0, +/-1.1)
c) (+/-2.5,0)
d) (+/-1.1,0)

Answer
1. Find and equation of the ellipse having the given points as foci and the given sum of the foci radii.
(-9,0); (9,0); 30
An ellipse has all the points equal distance from the same point.
Using this, the right x point must be at A, where the distance to both points is the same.
That must mean at the right end, X=A and Y=0, so (A+9)+(A-9)=30.

That is 2A=30, or A=±15.  Since the distance is 30, that means that where it crosses the y-axis is at (0,B) where 2√(9²+B²)=30.  Dividing by 2 gives √(9²+B²)=15.  Squaring both sides gives
81+B²=225.  Subtracting 81 gives B²=144, so B=±12.

The form of an ellipse is given by (x/A)² + (y/B)²=1.  Here, A and B were 15 and 12, repectively.

a)(x²/15)+ (y²/12)=1
b)(x²/225)+(y²/144)=1
c)(x²/15)-(y²/12)=1
d)(x²/225)-(y²/144)=1


2. Choose the correct coordinates of the foci: x^2+4y^2-16=0

This can be transformed into an ellipse is of the form (x/4)² + (y/2)² = 1.
We need 2√(A²+2²) = 4+A+ 4-A.  This is the same as √(A²+2²) = 4.
That means, after squaring both sides, that A²+4 = 16, so A=±√12.
Since the x is divided by the larger value, the foci must be on the x axis, so y=0.
That means that x = ±√12.

a)(12,0)and (-12,0)
b)(-2√3,0) and (2√3),0)
c)(-√12,0) and (√12,0)
d)(-4√3,0) and (4√3,0)


3. Find the equation of the hyperbola described:
Foci (0,-8) and (0,8); difference of focal radii 10

To be a hyperbola, there needs to be a minus sign, so that eliminate (b) and (d).
With a difference of 10, that means the foci are at ±5.
Since the equation has this value squared, it must be (a).


4. Find an equation of the hyperbola described:
Asymptotes y=x and y=-x; foci (-4,0) and (4,0)

To be a hyperbola, there needs to be a minus sign, so we're down to (a) or (c).
Since the foci are at x=±4, it must be (a).

a) (x²/8)-(y²2/8)=1
b) (x²/4)+(y^2/4)=1
c) x²-y²=1
d) x²+y²=1


5. Use the information to write an equation of the parabola:
Focus:(-8,-5), Vertex: (2,-5)

The equation of a parabola is (y−h)² = 4D(x−k) where D is the distance from the vertex
to the focus, and (h,k) is the vertex.  We are given h is -5, k is 2, and that D = 2+8 = 10.

a) (y+5)² = -40(x-2)
b) (x+5)² = 4(y-2)
c) (y-5)² = -4(x-2)
d) (x-5)² = 40(y-2)


6. Choose the correct foci:
5x²+9y²=45

Divide by 45 and get x²/9 + y²/5 = 1.
Since 9 > 5, the foci must be on the x axis.
Since there difference between them is the square of the foci position relative to the center,
that mean 9-5=4 is the square of the foci position relative to the center.
The √4 = ±2, so find ±2 on the x-axis, 0 on the y-axis.

a)(±2,0)
b)(±4,0)
c) (0,±2)
d) (0,±4)


7. Choose the correct foci: 9x² + 4y² = 9

The equation is x² + (2y/3)² = 1.
This puts the points on the axis at x = ±1 and y = ±3/2 = ±1.5.
Since 2.5 is greater than 1.5, it must be the 1.1.
Since the points on the axis are greater in the y direction, the foci must be on the y axis.
So, which one has ±1.1 on the y axis?

a) (0,±2.5)
b) (0, ±1.1)
c) (±2.5,0)
d) (±1.1,0)