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Question

Let $\mathrm{A}(\mathrm{a} \sec 	heta, \mathrm{btan} 	heta)$

tangent at $A$ is $\frac{x \sec 	heta}{a}-\frac{y 	an 	heta}{b}=1$

$\because$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Let $\mathrm{A}(\mathrm{a} \sec 	heta, \mathrm{btan} 	heta)$

tangent at $A$ is $\frac{x \sec 	heta}{a}-\frac{y 	an 	heta}{b}=1$

$\because$ tangent passes through $(0,-b) \Rightarrow 	an 	heta=1$

$	herefore A(a \sqrt{2}, b)$

$\Rightarrow \mathrm{CA} \cos 45^{\circ}=\mathrm{CM}$

$\Rightarrow \sqrt{2 a^{2}+4 b^{2}} \cdot \frac{1}{\sqrt{2}}=2 b$

$\Rightarrow \mathrm{a}^{2}=2 \mathrm{b}^{2}$

Let tangents drawn from point $C(0,-b)$ to hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ touches hyperbola at points $A$ and $B.$ If $\Delta ABC$ is a right angled triangle, then $\frac{a^2}{b^2}$ is equal to -

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