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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If the normal to the ellipse x ^ 2 25 + y ^ 2 1 = 1 is at a...

If the normal to the ellipse

\dfrac{{x}^{2}}{25} + \dfrac{{y}^{2}}{1} = 1

is at a distance

p

from the origin, then the maximum value of

p

is

Correct answer

4

Step-by-step solution

Equation of the normal is

\left(5sec \theta\right)x - \left(cosec\theta\right)y = 24

So,

p = \dfrac{24}{\sqrt{25\sec^{2} \theta + cose{c}^{2}\theta}} = \dfrac{24}{\sqrt{36 + {\left(5tan \theta - cot \theta\right)}^{2}}}

{p}_\max = \dfrac{24}{6} = 4

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