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 $
$m=-\frac{3}{2}$
$m=+\frac{3\sec \theta }{\sqrt{12} \cdot \tan \theta}$
$\Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2}$
$\sin \theta=-\frac{1}{\sqrt{3}}$
$(\sqrt{12} \cdot \sec \theta, 3 \tan \theta)$
$(\sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}},-3 \times \frac{1}{\sqrt{2}}) \Rightarrow(\frac{6}{\sqrt{2}}, \frac{-3}{\sqrt{2}})$
