- #1
moshee
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Homework Statement
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.
Homework Equations
(x-h)^2/b^2 + (y-k)^2/a^2 = k AND (y-k)^2=4a(x-h)
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.