### Hyperbola Question 5

#### Question 5 - 01 February - Shift 2

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 :

(1) -3

(2) 9

(3) -9

(4) 3

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Condition for tangency and points of contact, Parametric representation

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

$ 3 x+2 y=1 $

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

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

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

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

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

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