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Suppose a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ cuts the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ at points $P,Q$, and that the tangents to the second ellipse at $P,Q$ are perpendicular. Determine $\frac{a^2}{c^2}+\frac{b^2}{d^2}.$

I tried as: I take a point on inner ellipse and from that wrote the equation of tangent and then tried to find the intersection points to other ellipse . But this became too complicated.

Can anyone please give me how to proceed?

Semiclassical
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Koolman
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1 Answers1

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Try to prove first, then exploit, the following lemma:

Given the ellipse $E_1$ with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the locus of points $P$ such that the tangents to $E_1$ drawn through $P$ are perpendicular is a circle centered at the origin with radius $\sqrt{a^2+b^2}$. Additionally, if $P_1,P_2$ are the tangency points of the tangents from $P$, the envelope of the $P_1 P_2$ lines is an ellipse $E_2$ concentric and confocal with respect to $E$.

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Jack D'Aurizio
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  • Can you show me the calculations – Koolman Sep 07 '16 at 17:37
  • @koolman: it is way more instructive if you think about the given lemma and perform the computation by yourself. To learn mathematics is much more than putting the right symbols in the right place. I do not think that stating "the answer is one" really adds something to your knowledge or understanding. – Jack D'Aurizio Sep 07 '16 at 17:40
  • Based on my (abortive) attempts at this question, I came up with another one which you might find interesting: http://math.stackexchange.com/questions/1918247/locus-of-intersection-of-two-perpendicular-normals-to-an-ellipse. (Instead of tracking the point $P$, track the point $P'$ such that $PP_1P'P_2$ would be a rectangle.) – Semiclassical Sep 09 '16 at 15:57