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V0ODO0CH1LD

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## Homework Statement

Let [itex] k>0 [/itex] be such that [itex] (x^2-x)+k(y^2-y)=0 [/itex] defines an ellipse with focal length equal to [itex] 2 [/itex]. If [itex] (p,q) [/itex] are the coordinates of a point in the ellipse with [itex] q^2 - q\not=0 [/itex], then what is [itex] \frac{p-p^2}{q^2-q} [/itex]?

## Homework Equations

The fact that the sum of the distances from any point in an ellipse is equal to two times the length of the major axis.

## The Attempt at a Solution

First, I wrote the ellipse equation in "standard" form:

[tex] \frac{(x-\frac{1}{2})^2}{\frac{1+k}{4}}+\frac{(y-\frac{1}{2})^2}{\frac{1+k}{4k}}=1 [/tex]

Which means that the ellipse is centered at [itex] (\frac{1}{2},\frac{1}{2}) [/itex] and it's major axis is the x-axis. Because [itex] k [/itex] could never be less than [itex] 1 [/itex] since that would make the major axis be less than the specified focal length. And as far as I know, that is not allowed.

Anyway, I could solve for [itex] \frac{1+k}{4} [/itex] and [itex] \frac{1+k}{4k} [/itex] by using the fact that major axis of the ellipse squared is equal to the minor axis squared plus half the focal length squared, which gives:

[tex] (\frac{1+k}{4})^2= (\frac{1+k}{4k})^2+1 [/tex]

Then I could say that the sum of the distances from any point to each focus of the ellipse is always equal to [itex] 2(\frac{1+k}{4}) [/itex]. And that leaves the equation below to solve for [itex] \frac{p-p^2}{q^2-q} [/itex]:

[tex] \sqrt{(\frac{3}{2}+p)^2+q^2} + \sqrt{(\frac{5}{2}+p)^2+q^2}=2(\frac{1+k}{4}) [/tex]

Where [itex] k [/itex] is a known.

Okay, now I am definitely stuck.. So I have a few questions.

First; can I find the relationship of p and q in the expression [itex] \frac{p-p^2}{q^2-q} [/itex] just by knowing their relationship in [itex] \sqrt{(\frac{3}{2}+p)^2+q^2} + \sqrt{(\frac{5}{2}+p)^2+q^2} [/itex]?

Second; why is it that [itex] \frac{p-p^2}{q^2-q} [/itex] is a constant ratio for every point in the ellipse? I feel like that is some property of ellipses which if I knew it would make the problem way easier. Is that right?

Third; if I can't just do what I said above would I actually have to do something else? What?

Thanks!