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Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$.Then the square of distance of the point $(6,-4)$from the normal to the hyperbola $\alpha^2 x^2-9 y^2=9 \alpha^2$at its point $(\alpha-1, \alpha+2)$ is equal to $........$.
$116$
$115$
$114$
$113$
Solution
$\because P (3, \alpha)$ lies on $y^2=12 x$
$\Rightarrow \alpha= \pm 6$
But, $\left.\frac{ dy }{ dx }\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6$ reject $)$
Now, hyperbola $\frac{x^2}{9}-\frac{y^2}{36}=1$, normal at
$Q(\alpha-1, \alpha+2) \text { is } \frac{9 x}{5}+\frac{36 y}{8}=45$
$\Rightarrow 2 x +5 y -50=0$
Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to
$\left|\frac{2(6)-5(4)-50}{\sqrt{2^2+5^2}}\right|=\frac{58}{\sqrt{29}}$
$\Rightarrow \text { Square of distance }=116$
