(Polar Equations) Finding the equation of a ellipse and hyperbola

1. Given the vertices (2,0) and (10, pi), find the equation of the ellipse.
2. Given the vertices (1, 3pi/2) and (9, 3pi/2), find the equation of the hyperbola.
I seriously have no clue on what do to for these type of problems.

ellipsis should probably be something like this:

(sin(theta)/A)²+(cos(theta)/B)²=1/r²
With A and B being the maximum distance they go.

from the cartesian equation:
(x/A)²+(y/B)²=1


But I think with the vertices you should build the equations from the ellipsis definition:
"A cord of fixed length tied to the two vertices will pass by all points of an ellipsis when held stretched"

 Spoiler

I seriously have no clue on what do to for these type of problems.

That's because there are many solutions to these equations. Take the first one for example:

1. Given the vertices (2,0) and (10, pi), find the equation of the ellipse.

The standard form of an ellipse is x^2/(r1)^2+y^2/(r2)^2=1, where r1 is the distance from the center to the x-vertex and r2 is the distance from the center to the y-vertex.

So as you can see, there are three possibilities for these two points:

1. The two points are the left vertex and the top vertex of an un-rotated ellipse.

The center of the ellipse is at (10,0), and the ellipse has an x-radius (r1) of 8 and a y-radius (r2) of π. Therefore, the equation of the ellipse is:

(x-10)^2/64+y^2/π^2=1

The x-10 is to shift the ellipse to a new center.

2. The two points are the bottom vertex and the right vertex of an un-rotated ellipse.

The center of the ellipse is at (2,π, and the ellipse has an x-radius of 8 and a y-radius of π. Therefore, using the same logic as 2:

(x-2)^2/64+(y-π^2/π^2=1

3. If the two points are opposite ends of a rotated ellipse, the points can yield infinitely many solutions depending on the unknown radius. I'm not going to try this right now because I don't remember how to do rotated conics, but the general process is:

Notice that the x-radius (r1) is:

sqrt((10-2)^2+π^2)/2 = sqrt(64+π^2)/2

The center is (6,π/2). Thus, "unrotated" form of the equation is:

4*(x-6)^2/(64+π^2)+(y-π/2)^2/(r2)^2=1

Then you find that the angle of rotation is:

φ = arctan(π/(10-2)) = arctan(π/8)

You take this, plug it into:

x = x'cos(φ - y'sin(φ
y = x'sin(φ + y'cos(φ

Substitute x and y and solve for x' and y'. The result is the rotated equation. Then specify that r1 can be any constant.

2. Given the vertices (1, 3pi/2) and (9, 3pi/2), find the equation of the hyperbola.

This can yield infinitely many solutions depending on the eccentricity of the hyperbola. The general form of a (sideways) hyperbola is:

x^2/(r1)^2-y^2/(r2)^2=1

Notice that r1 is 4, and the center is at (5,3π/2) Therefore:

(x-5)^2/16-(y-3π/2)^2/(r2)^2=1

r2 can be any constant except for 0.

-------

Edit: Crap. I didn't see that this is in polar form. That drastically simplifies things. For polar equation, the origin is always one of the focal points.

The general equation for a horizontal ellipse that has the center at π is:

r(θ = a*(1-e^2)/(1+e*cos(θ)

The general equation for a vertical hyperbola that has center at 3π/2 is:

r(θ = a*(e^2-1)/(1+e*sin(θ)

e is the eccentricity.

e = c/a

where c is the distance from the center of the conic to the focus
and a is the radius along the longer axis

This is the same a as the a in the equation.

Plug and chug. I wasted so much time doing it in rectangular. Lol.

If you want to take an hour or two and really learn it, I bet Khan can help.

http://www.khanacademy.org/video/introdu...st=Algebra
That's the starting point, and there are about 10 videos of relevance following that. If you make an account, there are exercises as well.