A(a, 0) and B(-a, 0) are two fixed points. If | vedclass

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(D) Let $C = (h, k)$.In $	riangle CDA$,$	an A = \frac{k}{a-h}$.In $	riangle CDB$,$	an B = \frac{k}{h-(-a)} = \frac{k}{h+a}$.Given $\angle A - \ang

Vedclass · Accepted Answer

$x^2 - y^2 + 2xy \cot 	heta = a^2$

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(D) Let $C = (h, k)$.In $	riangle CDA$,$	an A = \frac{k}{a-h}$.In $	riangle CDB$,$	an B = \frac{k}{h-(-a)} = \frac{k}{h+a}$.Given $\angle A - \angle B = 	heta$,so $	an(A - B) = 	an 	heta$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = 	an 	heta$,$\frac{\frac{k}{a-h} - \frac{k}{h+a}}{1 + \frac{k}{a-h} \cdot \frac{k}{h+a}} = 	an 	heta$.$\frac{k(h+a) - k(a-h)}{(a-h)(h+a) + k^2} = 	an 	heta$.$\frac{2kh}{a^2 - h^2 + k^2} = 	an 	heta$.$2kh = (a^2 - h^2 + k^2) 	an 	heta$.$2kh \cot 	heta = a^2 - h^2 + k^2$.$h^2 - k^2 + 2hk \cot 	heta = a^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 + 2xy \cot 	heta = a^2$.

$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

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