Let $(x_1,y_1)$ be the point at which the is $||$ to $ 2x+3y=6$

Therefore slope of the at $(x_1,y_1)$ is $m=\large\frac {-2}{3}$ (slope of the line)

Step 2:

$m=\large\frac{dy}{dx_{(x_1y_1)}}$

Differentiating the equation with respect to x

$2x+2y \large\frac{dy}{dx}$$=0$

Step 3:

$\large\frac{dy}{dx_{(x_1,y_1)}}=\large\frac{-x_1}{y_1}=\frac{-2}{3}$

=> $\large\frac{x_1}{\sqrt {52-{x_1}^2}}=\frac{2}{3}$

=>$3x_1=2 \sqrt{52-{x_1}^2}$

$9{x_1}^2=208-4{x_1}^2$

Therefore $13 {x_1}^2=208=>{x_1}^2=16=>x_1=\pm 6$

$y_1=\sqrt {52-x_1^2}=\sqrt {52-36}=\pm 4$

Step 4:

$x_1,y_1$ are of the same sign since $ \large\frac{x_1}{y_1} $$>0$

Therefore the points are (6,4),(-6,-4)

Step 5:

Equation of the tangent at (6,4)

$y-4=\large\frac{-2}{3}$$(x-6)$

$3y+12=-2x-12$

$2x+3y+24=0$

The tangent are $2x+3y\pm 24=0$