Step 1:
Tangent and normal at $(\sqrt 3,0)$ to the ellipse $2x^2+3y^2=6$
The tangent at $(x_1,y_1)$ is $2xx_1+3yy_1=6$
Here $(x_1,y_1)=(\sqrt 3,0)$
The tangent is $2\sqrt 3 x=6$
$\Rightarrow \sqrt 3x-3=0$
$x-\sqrt 3=0$
Step 2:
The normal $\perp$ to the tangent and is of the form $y=k$.It passes through $(\sqrt 3,0)\Rightarrow k=0$.
Normal is $y=0$
Equation of tangent $x=\sqrt 3$
Equation of normal $y=0$
Note : $(\sqrt 3,0)$ can be seen to be one of the vertices of the ellipse.The tangent at the vertex $(a,0)$ is $x=a$.So the tangent is $x=\sqrt 3$).