If the normal to the ellipse x^225+y^21=1 is at a distance p from the origin, then the maximum value of p is

Correct answer: 4. Equation of the normal is 5sec⁡θx-cosecθy=24 So, p=24/(√(25sec^2⁡θ+cosec^2θ))=24/(√(36+5tan⁡θ-cot⁡θ^2)) p_max=24/6=4

Q4 / 53 · NTA Abhyas (JEE Main) 2020 · Mathematics · Ellipse

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

Correct answer: 4

Equation of the normal is

(5 \sec θ) x - (c o s e c θ) y = 24

So,

p = \frac{24}{\sqrt{25 \sec^{2} θ + c o s e c^{2} θ}} = \frac{24}{\sqrt{36 + {(5 \tan θ - \cot θ)}^{2}}}

p_{m a x} = \frac{24}{6} = 4

Hence, the correct answer is 4.