Let the tangent to the parabola y^2=12 x at t | vedclass

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Question

$\because P (3, \alpha)$ lies on $y^2=12 x$

$\Rightarrow \alpha= \pm 6$

But, $\left.\frac{ dy }{ dx }\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \R

Vedclass · Accepted Answer

$116$

Vedclass · Answer

$\because P (3, \alpha)$ lies on $y^2=12 x$

$\Rightarrow \alpha= \pm 6$

But, $\left.\frac{ dy }{ dx }ight|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6$ reject $)$

Now, hyperbola $\frac{x^2}{9}-\frac{y^2}{36}=1$, normal at

$Q(\alpha-1, \alpha+2) 	ext { is } \frac{9 x}{5}+\frac{36 y}{8}=45$

$\Rightarrow 2 x +5 y -50=0$

Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to

$\left|\frac{2(6)-5(4)-50}{\sqrt{2^2+5^2}}ight|=\frac{58}{\sqrt{29}}$

$\Rightarrow 	ext { Square of distance }=116$

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 $........$.

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