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

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(D) Let the coordinates of point $C$ be $(h, k)$.Given $\angle A - \angle B = 	heta$,so $	an(A - B) = 	an 	heta$ .....$(i)$In the right-angled tri

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

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(D) Let the coordinates of point $C$ be $(h, k)$.Given $\angle A - \angle B = 	heta$,so $	an(A - B) = 	an 	heta$ .....$(i)$In the right-angled triangle $CDA$,where $D$ is the projection of $C$ on the $x$-axis,$	an A = \frac{k}{a - h}$.In the right-angled triangle $CDB$,$	an B = \frac{k}{h - (-a)} = \frac{k}{h + a}$.From $(i)$,we have $\frac{	an A - 	an B}{1 + 	an A 	an B} = 	an 	heta$.Substituting the values of $	an A$ and $	an B$:$\frac{\frac{k}{a - h} - \frac{k}{a + h}}{1 + \frac{k^2}{a^2 - h^2}} = 	an 	heta$$\frac{k(a + h) - k(a - h)}{a^2 - h^2 + k^2} = 	an 	heta$$\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 ${x^2} - {y^2} + 2xy \cot 	heta = {a^2}$.

Two fixed points are $A(a, 0)$ and $B(-a, 0)$. If $\angle A - \angle B = \theta$,then the locus of point $C$ of triangle $ABC$ will be

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