Question

Let $P\left(x_0,y_0\right)$ be the point on the hyperbola $3x^2-4y^2=36$, which is nearest to the line $3x+2y=1$. Then $\sqrt{2}\left(y_0-x_0\right)$ is equal to :

• A. $-9$
• B. $3$
• C. $9$
• D. $-3$

Answer 1

Answer: (A)

$3x^2-4y^2=363x+2y=1$
$m=-\frac{3}{2}$
$m=+\frac{\sec \theta 3}{\sqrt{12}\cdot \tan \theta }$
$\Rightarrow \frac{3}{\sqrt{12}}\times \frac{1}{\sin \theta }=\frac{-3}{2}$
$\sin \theta =-\frac{1}{\sqrt{3}}$
$(\sqrt{12}\cdot \sec \theta ,3\tan \theta )$
$\left(\sqrt{12}\cdot \frac{\sqrt{3}}{\sqrt{2}},-3\times \frac{1}{\sqrt{2}}\right)\Rightarrow \left(\frac{6}{\sqrt{2}},\frac{-3}{\sqrt{2}}\right)$