Let \mathrm{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 :

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Answer: B. -9 If \left(x_0, y_0ight) is point on hyperbola then tangent at \left(x_0, y_0ight) is parallel to 3 x+2 y=1 Equation of tangent = \frac{x x_0}{12}-\frac{y y_0}{9}=2 Slope of tangent =\frac{-3}{2} Equation of tangent in slope form y=\frac{-3}{2} x \pm \sqrt{12 \cdot \frac{9}{4}-9} y=\frac{-3}{2} x \pm 3 \sqrt{2} or 3 x+2 y=6 \sqrt{2} Comparing \begin{aligned} & \frac{\frac{x_0}{12}}{3}=\frac{\frac{-y_0}{9}}{2}=\frac{1}{6 \sqrt{2}} \\ & x_0=3 \sqrt{2}, y_0=\frac{-3}{\sqrt{2}} \\ & \sqrt{2}\left(y_0-x_0ight)=-3-6=-9 \\ & \end{aligned}

Let $$\mathrm{P}\left(x_{0}, y_{0}\right)$$ 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 :

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