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

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