Let P ( x 0 , y 0 ) be the point on the hyperbola 3 x 2 - 4 y 2 = 36 , which is nearest to the line 3 x + 2 y = 1 . Then 2 ? ( y 0 - x 0 ) is equal to [2023]

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Solution

Q.
Let $P (x_{0}, y_{0})$ be the point on the hyperbola $3 x^{2} - 4 y^{2} = 36$ , which is nearest to the line $3 x + 2 y = 1 .$ Then $\sqrt{2} (y_{0} - x_{0})$ is equal to [2023]

1 - 9

2 3

3 9

4 - 3

Ans.
(1)

We have, $3 x^{2} - 4 y^{2} = 36$

$The slope of 3 x + 2 y = 1 is$ $m = - \frac{3}{2}$

$The point of hyperbola is (\sqrt{12} \sec θ, 3 \tan θ) .$

$The equation of normal is$ $\sqrt{12} x \cos θ + 3 y \cot θ = 21$

$\Rightarrow Slope$ $= - \frac{\sqrt{12} \cos θ}{3 \cot θ}$

$According to questions,$ $(- \frac{3}{2}) (\frac{- \sqrt{12} \cos θ}{3 \cot θ}) = - 1$

$\Rightarrow \sqrt{3} \sin θ = - 1$ $\Rightarrow \sin θ = - \frac{1}{\sqrt{3}}$ $\Rightarrow \cos θ = \sqrt{\frac{2}{3}}$

$∴$ $(x_{0}, y_{0}) \equiv (\frac{6}{\sqrt{2}}, - \frac{3}{\sqrt{2}})$ $∴$ $\sqrt{2} (y_{0} - x_{0}) = - 9$

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