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

Answer: 4

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