Question

The coordinates of a point on the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$ which is nearest to the line $3x+2y+1=0$ are

• A. $(6,3)$
• B. $(-6,-3)$
• C. $(6,-3)$
• D. $(-6,3)$

Answer 1

Answer: (C)

Point $P$ is nearest to the given line if the tangent at $P$ is parallel to the given line.

Now, the slope of tangent at $P\left(x_1,y_1\right)$ is
${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{18y_1}{24x_1}=\frac{3}{4}\frac{y_1}{x_1}$
which must be equal to $-3/2$.
Therefore,
$\frac{3}{4}\frac{y_1}{x_1}=-\frac{3}{2}$
or $y_1=-2x_1$.... (i)
Also, $\left(x_1,y_1\right)$ lies on the curve.
Hence, $\frac{x_1^2}{24}-\frac{y_1^2}{18}=1$ (ii)
Solving (i) and (ii), we get two points $(6,-3)$ and $(-6,3)$ of which $(6,-3)$ is the nearest.