MCQ
Let $P \left( x _0, y _0\right)$ 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 $\sqrt{2}\left( y _0- x _0\right)$ is equal to :
  • A
    $-3$
  • B
    $9$
  • $-9$
  • D
    $3$

Answer

Correct option: C.
$-9$
c
$3 x^2-4 y^2=36 \quad 3 x+2 y=1$

$m =-\frac{3}{2}$

$m =+\frac{\sec \theta 3}{\sqrt{12} \cdot \tan \theta}$

$\Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2}$

$\sin \theta=-\frac{1}{\sqrt{3}}$

$(\sqrt{12} \cdot \sec \theta, 3 \tan \theta)$

$\left(\sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}},-3 \times \frac{1}{\sqrt{2}}\right) \Rightarrow\left(\frac{6}{\sqrt{2}}, \frac{-3}{\sqrt{2}}\right)$

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