If L is the length of the perpendicular drawn from the origin to any normal of the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1$ then the maximum value of L is

$$\begin{gathered} Given ellipse is \frac{x^2}{25}+\frac{y^2}{16}=1(a>b) \\ where a^2=25,b^2=16 \\ Let P\left(\theta \right)=(5sec\theta ,4\cos ec\theta ) be any point on ellipse \end{gathered}$$

Equation of normal to ellipse at P  is 

$\begin{array}{l}5\sec \mathrm{\theta x}-4\mathrm{cosec}\mathrm{\theta y}-9=0\\ \mathrm{L}=\frac{9}{\sqrt{25{\sec }^2\mathrm{\theta }+16{\mathrm{cosec}}^2\mathrm{\theta }}}\\ =\frac{9}{\sqrt{25{\tan }^2\mathrm{\theta }+16{\cot }^2\mathrm{\theta }+41}}\\ For denominator apply A.M\ge G.M\\ \sqrt{25{\tan }^2\mathrm{\theta }+16{\cot }^2\mathrm{\theta }+41}\ge \sqrt{81}\\ \Rightarrow \frac{9}{\sqrt{25{\tan }^2\mathrm{\theta }+16{\cot }^2\mathrm{\theta }+41}}\le 1\\ \Rightarrow \mathrm{L}\le 1\\ Maximum value of L is 1\end{array}$