Imagine Wrote: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:

Imagine Wrote: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.

Imagine Wrote: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.