Let [P(x_0, y_0)] be the point on the hyperbola 3 x^2-4 y^2=36, which is nearest to the li…

Question

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 :

Accepted Answer

If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$

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

$$
\begin{aligned}
& \frac{\frac{x_0}{12}}{3}=\frac{\frac{-y_0}{9}}{2}=\frac{1}{6 \sqrt{2}} \\
& x_0=3 \sqrt{2}, y_0=\frac{-3}{\sqrt{2}} \\
& \sqrt{2}\left(y_0-x_0\right)=-3-6=-9 \\
&
\end{aligned}
$$

Solutions · 1