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The locus of a point whose chord of contact w.r.t. the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ touches the circle described on the straight line joining the foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as diameter is
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$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}$
Circle on the join of foci $(a e, 0)$ and $(-a e, 0)$ diameter is:
$\begin{aligned} & \Rightarrow(x-a e)(x+a e)+(y-0)(y-0)=0 \\ & \Rightarrow x^2+y^2=a^2 e^2=a^2+b^2 \text { WC...(1) } \\ & {\left[a^2 e^2=a^2+b^2\right]}\end{aligned}$
Let chord of contact of $P\left(x_1, y_1\right)$ touch the circle (1)
Equation of chord of contact $P$ is $[T=0]$
$\begin{aligned} & \Rightarrow \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1 \quad \ldots(2) \\ & \Rightarrow b^2 x_1 x-a^2 y_1 y-a^2 b^2=0 \\ & \Rightarrow \frac{a^3 b^2}{\sqrt{\left(b^4 x_1^2+a^4 y_1^2\right)}}= \pm \sqrt{\left(a^2+b^2\right)}\end{aligned}$
Hence locus of $P\left(x_1, y_1\right)$ is $\left(b^4 x^2+a^4 y^2\right)\left(a^2+b^2\right)=a^4 b^4$.
$\begin{aligned} & \Rightarrow(x-a e)(x+a e)+(y-0)(y-0)=0 \\ & \Rightarrow x^2+y^2=a^2 e^2=a^2+b^2 \text { WC...(1) } \\ & {\left[a^2 e^2=a^2+b^2\right]}\end{aligned}$
Let chord of contact of $P\left(x_1, y_1\right)$ touch the circle (1)
Equation of chord of contact $P$ is $[T=0]$
$\begin{aligned} & \Rightarrow \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1 \quad \ldots(2) \\ & \Rightarrow b^2 x_1 x-a^2 y_1 y-a^2 b^2=0 \\ & \Rightarrow \frac{a^3 b^2}{\sqrt{\left(b^4 x_1^2+a^4 y_1^2\right)}}= \pm \sqrt{\left(a^2+b^2\right)}\end{aligned}$
Hence locus of $P\left(x_1, y_1\right)$ is $\left(b^4 x^2+a^4 y^2\right)\left(a^2+b^2\right)=a^4 b^4$.
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