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Question

A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is
x2a2-y2b2=1, where b2 = a2 (e2 − 1).

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Solution

Let A-ae,0 and Bae,0 be the given points. Let P(h, k) be a point such that PA-PB=2a.

h+ae2+k-02-h-ae2+k-02=2ah+ae2+k2=2a+h-ae2+k2

Squaring both sides, we get:

h2+a2e2+2aeh+k2=4a2+h2+a2e2-2aeh+k2+4ah-ae2+k2aeh=2a2-aeh+2ah-ae2+k2eh-a=h-ae2+k2 a0

Squaring both sides again, we get:

e2h2+a2-2aeh=h2+a2e2-2aeh+k2 e2h2+a2=h2+a2e2+k2 a2e2-1=h2e2-1-k2h2a2-k2a2e2-1=1

Hence, the locus of (h, k) is x2a2-y2b2=1, where b2=a2e2-1.

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