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 :
Solution
If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$
<br/><br/>Equation of tangent $= \frac{x x_0}{12}-\frac{y y_0}{9}=2$
<br/><br/>Slope of tangent $=\frac{-3}{2}$
<br/><br/>Equation of tangent in slope form
<br/><br/>$y=\frac{-3}{2} x \pm \sqrt{12 \cdot \frac{9}{4}-9}$
<br/><br/>$y=\frac{-3}{2} x \pm 3 \sqrt{2}$
<br/><br/>or $3 x+2 y=6 \sqrt{2}$
<br/><br/>Comparing
<br/><br/>$$
\begin{aligned}
& \frac{\frac{x_0}{12}}{3}=\frac{\frac{-y_0}{9}}{2}=\frac{1}{6 \sqrt{2}} \\\\
& x_0=3 \sqrt{2}, y_0=\frac{-3}{\sqrt{2}} \\\\
& \sqrt{2}\left(y_0-x_0\right)=-3-6=-9 \\\\
&
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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