Find the point an curve $3x^2-4y^2=72$ which is nearest to the line $3x+2y+1=0\mathrm{is}$

Slope of line $3x+2y+1=0is$ $\frac{-3}{2}$

Diff  the curve  $3x^2-4y^2=72$ we get 

$\frac{dy}{dx}=\frac{3x}{4y}$

$m={\left(\frac{dy}{dx}\right)}_{p(x_1,y_1)}=\frac{3}{4}\frac{x_1}{y_1}$               $\frac{-3}{2}=\frac{3}{4}\frac{x_1}{y_1}$        $\frac{x_1}{y_1}=-2$

$p(x_1,y_1)liesan3x^2-4y^2\Rightarrow 72$                          $3{x_1}^2-4{y_1}^2\Rightarrow 72$

$3\frac{{x_1}^2}{{y_1}^2}-4=\frac{72}{{y_1}^2}$        $3\left(4\right)-4=\frac{72}{{y_1}^2}$              $y_1=\pm 3$

Rev . points are (-6,3) to (6,-3)

    Distance of (-6,3) to  $3x+2y+1=0$ is $\frac{\left|-18+6+1\right|}{\sqrt{9+4}}=\frac{11}{\sqrt{13}}$

                     Distance of (6,-3) to $3x+2y+1=0$ is           $\sqrt{13}$      (-6,3)