The normal to the ellipse $(x^2/9) + (y^2/4) = 1$ which is farthest from its center is  .

$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$

$\perp$ distance from origin

$P=\left|\frac{a^2-b^2}{\sqrt{a^{2}se{c}^2\theta +b^2\cos e{c}^2\theta }}\right|$

for $P_{max}$ $Dr$ = min.

                               $$\begin{gathered} =\sqrt{{(a \tan \theta - b cot\theta )}^2+a^2+b^2+2ab} \\ =\sqrt{{(a \tan \theta - b cot\theta )}^2+{(a+b)}^2} \\ =a+b \end{gathered}$$

so                             $P_{max}=\frac{a^2-b^2}{a+b}=(a-b)$

at                              $\frac{b}{a}={\tan }^2\theta \Rightarrow \tan \theta =\sqrt{\frac{b}{a}}$

so normal are           $$\begin{gathered} \pm \frac{ax\sqrt{a+b}}{\sqrt{a}}\pm \frac{by\sqrt{a+b}}{\sqrt{b}}=a^2-b^2 \\ \pm \sqrt{3} \sqrt{5}x \pm \sqrt{2} \sqrt{5} y =5 \end{gathered}$$

i.e.                              $\pm \sqrt{3} x \pm \sqrt{2 }y =\sqrt{5}$