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10-2. Parabola, Ellipse, Hyperbola
hard
माना अतिपरवलय $3 \mathrm{x}^2-4 \mathrm{y}^2=36$ पर बिन्दु $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$, रेखा $3 \mathrm{x}+2 \mathrm{y}=1$ के निकटतम है। तो $\sqrt{2}\left(\mathrm{y}_0-\mathrm{x}_0\right)$ बराबर है:
A
$-3$
B
$9$
C
$-9$
D
$3$
(JEE MAIN-2023)
Solution
$3 x^2-4 y^2=36 \quad 3 x+2 y=1$
$m =-\frac{3}{2}$
$m =+\frac{\sec \theta 3}{\sqrt{12} \cdot \tan \theta}$
$\Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2}$
$\sin \theta=-\frac{1}{\sqrt{3}}$
$(\sqrt{12} \cdot \sec \theta, 3 \tan \theta)$
$\left(\sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}},-3 \times \frac{1}{\sqrt{2}}\right) \Rightarrow\left(\frac{6}{\sqrt{2}}, \frac{-3}{\sqrt{2}}\right)$
Standard 11
Mathematics
