The coordinate of the point $p\left(x,y\right)$  lying in the first quadrant on the ellipse

$\frac{x^{\text{2}}}{8}\text{+}\frac{{\text{y}}^{\text{2}}}{18}=1$  so that the area of the triangle formed by the tangent at p and the coordinate axes is the smallest , are given by

 

Any point on the ellipse is given by $\left(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta \right)$

Now      $\frac{2x}{8}+\frac{2}{18}y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{9x}{4y}$

$\Rightarrow \frac{dy}{dx}|{}_{\left(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta \right)}=-\frac{9\sqrt{8}\cos \theta }{4\sqrt{18}\sin \theta }=-\frac{\sqrt{9}}{2}\cot \theta$

Hence the equation of the tangent at $\left(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta \right)$  is

$y-\sqrt{18}\sin \theta =-\frac{\sqrt{9}}{2}\cot \theta \left(x-\sqrt{8}\cos \theta \right)$

Therefore , the tangent cuts the coordinate axes at the points

$\left(0,\frac{\sqrt{18}}{\sin \theta }\right)and\left(\frac{\sqrt{8}}{\cos \theta },0\right)$

Thus the area of the triangle formed by this tangent and the coordinate axes is

$A=\frac{1}{2}\sqrt{18}.\sqrt{8}.\frac{1}{\cos \theta \sin \theta }$

     $=\frac{6}{\cos \theta \sin \theta }=12\cos ec2\theta$

But $\cos ec2\theta$  is smallest when $\theta =\pi /4.$

Therefore A is smallest when $\theta =\pi /4.$

Hence the required point is $\left(\sqrt{8}.\frac{1}{\sqrt{2}},\sqrt{18}.\frac{1}{\sqrt{2}}\right)=\left(2,3\right)$