Conics Assignment: prove square inscribed in the ellipse b^2x^2+a^2b^2 is equal...

[mathgirl10]

Prove that the area of a square inscribed in the ellipse b^2x^2+a^2b^2 is equal 4a^2b^2/a^2+b^2

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[ksdhart2]

Erm... sorry, but this all makes no sense to me. I honestly haven't even the slightest clue what any of it means. You say "b^2x^2+a^2b^2" but that's neither an ellipse, nor a "square inscribed in [an] ellipse."

 

[HallsofIvy]

Prove that the area of a square inscribed in the ellipse b^2x^2+a^2b^2 is equal 4a^2b^2/a^2+b^2

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Do you mean \(\displaystyle b^2x^2+ a^2y^2= a^2b^2\)?
Then for any x, \(\displaystyle y= \frac{b}{a}\sqrt{a^2- x^2}\).

In that case, a rectangle, with horizontal side lengths s, has vertices at \(\displaystyle \left(\frac{s}{2}, \frac{b}{a}\sqrt{a^2- \frac{s^2}{4}}\right)\), \(\displaystyle \left(-\frac{s}{2}, \frac{b}{a}\sqrt{a^2- \frac{s^2}{4}}\right)\), \(\displaystyle \left(-\frac{s}{2}, -\frac{b}{a}\sqrt{a^2- \frac{s^2}{4}}\right)\), and \(\displaystyle \left(\frac{s}{2}, -\frac{b}{a}\sqrt{a^2- \frac{s^2}{4}}\right)\).

The vertical side lengths are \(\displaystyle \frac{2b}{a}\sqrt{a^2- \frac{s^2}{4}}\) and, since this a square that must be equal to s.

Find s such that \(\displaystyle \frac{2b}{a}\sqrt{a^2- \frac{s^2}{4}}= s\), in terms of a and b, then find the area of that square.