Question

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. $-9$
• B. $3$
• C. $9$
• D. $-3$

Answer 1

Answer: (A)

$3 x ^2-4 y ^2=36 \,\,\,\, 3 x +2 y =1$
$ 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)$