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  • Let the tangent to the parabola at the point <img alt="(3, \alpha)" src="h

Let the tangent to the parabola y^{2}=12 \mathrm{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______.

Option: 1

116


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

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

Now, hyperbola\; \; \frac{x^{2}}{9}-\frac{y^{2}}{36}=1, normal \; at
\mathrm{Q}(\alpha-1, \alpha+2) \text { is } \frac{9 \mathrm{x}}{5}+\frac{36 \mathrm{y}}{8}=45

\Rightarrow 2 \mathrm{x}+5 \mathrm{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$ Square of distance $=116

Posted by

SANGALDEEP SINGH

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