The line $x \cos \alpha+y \sin \alpha=p$ touches the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if
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The line $x \cos \alpha+y \sin \alpha=p$ touches the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if

(A) $a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$

(B) $a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=\dot{p}$

(C) $a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2 c$

(D) $a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p$

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SOLUTION —

The line $x \cos \alpha+y \sin \alpha=p$ touches the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text {, }$ then

$c^2=a^2 m^2-b^2$

$\begin{array}{rlrl}\Rightarrow \left(\frac{p}{\sin \alpha}\right)^2  =a^2\left(\frac{\cos \alpha}{\sin \alpha}\right)^2-b^2 \\\Rightarrow  p^2  =a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha\end{array}$

So, The correct option will be (A).

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