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The line $x \cos \alpha+y \sin \alpha=p$ will touch the parabola $y^2=4 a(x+a)$, if
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Verified Answer
The correct answer is:
$p \cos \alpha+a=0$
$x \cos \alpha+y \sin \alpha-p=0 \ldots(i)$
$2 a x-y y_1+2 a\left(x_1+2 a\right)=0 \ldots(ii)$
From (i) and (ii), $\frac{\cos \alpha}{2 a}=\frac{\sin \alpha}{-y}=\frac{-p}{2 a(x+2 a)}$
$\Rightarrow y=-2 a \tan \alpha$ and $x=-p \sec \alpha-2 a$
$\therefore y^2=4 a(x+a) \Rightarrow 4 a^2 \tan ^2 \alpha=-4 a(p \sec \alpha+a)$
$\Rightarrow p \cos \alpha+a=0$
$2 a x-y y_1+2 a\left(x_1+2 a\right)=0 \ldots(ii)$
From (i) and (ii), $\frac{\cos \alpha}{2 a}=\frac{\sin \alpha}{-y}=\frac{-p}{2 a(x+2 a)}$
$\Rightarrow y=-2 a \tan \alpha$ and $x=-p \sec \alpha-2 a$
$\therefore y^2=4 a(x+a) \Rightarrow 4 a^2 \tan ^2 \alpha=-4 a(p \sec \alpha+a)$
$\Rightarrow p \cos \alpha+a=0$
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