# Determining the equation of an ellipse from its intersection with a parabola

1. Jun 8, 2009

### moshee

1. The problem statement, all variables and given/known data

The vertex of the parabola y^2=2px is the center of an ellipse. The focus of the parabola is an end of one of the principle axes of the ellipse, and the parabola and ellipse intersect at right angles. Find the equation of the ellipse.

2. Relevant equations

(x-h)^2/b^2 + (y-k)^2/a^2 = k AND (y-k)^2=4a(x-h)

3. The attempt at a solution

Here is my thought process and what I have obtained thus far:

y^2=2px is of the general form (y-k)^2=4a(x-h) i.e. a parabola whose directrix is parallel to the y-axis. The vertex is at (0,0) which also implies that the center of the ellipse is at (0,0) by the problem definition. The focus of the parabola is simply (p/2,0) and thus the coordinates of the endpoints for one of the principle axes (major or minor) is (+/- p/2,0).

Because the ellipse and parabola intersect at right angles, I have constructed my ellipse as: (x^2/b^2) + (y^2/a^2) = 1 (a>b) so that the foci which lie on the major axis of the ellipse are at (0,+/- c).

Substituting what I have so far into the eq'n of the ellipse will yield:
(4x^2)/p^2 + (y^2/a^2) = 1

To solve for a^2 and b^2, I applied the two basic properties of the ellipse. (1) The sum of the distances from any point on the ellipse to the foci is equal to the length of the major axis of the elipse and (2) c^2 = a^2-b^2 (where a & b are major/minor axis respectively)

Applying the first property I use the point (p/2,0) and establish the equality:

sqrt[(p/2-0)^2 + (0-c)^2] + sqrt[(p/2-0)^2 + (0+c)^2] = 2a
Simplifying: p^2+4c^2=4a^2

Applying the second property I have: c^2 = a^2 - (p/2)^2

2 eqn's, 2 unknowns. However, solving the equations i get 0=0, which is clearly wrong. I am not sure where I went wrong and need help uncovering my error. The suggested answer to the problem is 4x^2+2y^2=p^2.

2. Jun 8, 2009

### tiny-tim

Welcome to PF!

Hi moshee! Welcome to PF!

(try using the X2 tag just above the Reply box )
This is a bit complicated.

Easier to find where they intersect, and then find the gradients at that point.

3. Jun 9, 2009

### moshee

Thanks for the two tips tiny-tim. I was able to solve for a2 using your idea. Since the parabola and ellipse intersect at right angles, the product of the two tangents at their intersection is -1. I first solved for y2 using
(dy/dx)ellipse x (dy/dx)parabola = -1, then equated this to the equation of the parabola, y2=2px.

Once again, thanks.