Question

If the normal to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{1}=1$ is at a distance $p$ from the origin, then the maximum value of $p$ is

Answer 1

Equation of the normal is
$\left(5 sec \theta \right)x-\left(c o s e c \theta \right)y=24$
So, $p=\frac{24}{\sqrt{25 \left(sec\right)^{2} \theta + c o s e c^{2} \theta }}=\frac{24}{\sqrt{36 + \left(5 tan ⁡ \theta - cot ⁡ \theta \right)^{2}}}$
$p_{m a x}=\frac{24}{6}=4$