Question

What is the maximum distance of the normal drawn from a variable point of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ from the centre of the ellipse.

• A. $2$ if $a=5,b=3$
• B. $10$ if $a=5,b=3$
• C. $2$ if $a=6,b=4$
• D. $10$ if $a=6,b=4$

Answer 1

Answer: (A), (C)

$2$ if $a=6,b=4$

Let $P(a\cos \theta ,b\sin \theta )$ be a variable point on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Therefore, $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$
$\frac{dy}{dx}{|}_p=-\frac{b}{a}\frac{x}{y}=-\frac{b}{a}\cot \theta$
$\therefore$ Equation of normal at $P$
$\Rightarrow y-b\sin \theta =\frac{-1}{\frac{b}{a}\cot \theta }(x-a\cos \theta )$
$\Rightarrow y-b\sin \theta =\frac{a}{b}\tan \theta \cdot (x-a\cos \theta )$
$\Rightarrow ax\sin \theta -a^2\sin \theta \cos \theta =by\cos \theta -b^2\sin \theta \cos \theta$
$\Rightarrow ax\sin \theta -by\cos \theta =(a^2-b^2)\sin \theta \cos \theta$
$\Rightarrow ax\sec \theta -by\text{cosec}\theta =(a^2-b^2)$
Centre of standard ellipse is origin.
Length of perpendicular from ${}^{\prime }{O}^{\prime }$ on the normal at ${}^{\prime }{P}^{\prime }$ is
$l=\frac{a^2-b^2}{\sqrt{a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta }}.....\left(i\right)$
For $l$ to be maximum then denominator should be minimum.
Let $Z=a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta ...\left(ii\right)$
So, $\frac{dZ}{d\theta }=0\Rightarrow 2a^2{\sec }^2\theta \tan \theta +2b^2{\text{cosec}}^2\theta \cot \theta =0$
$\therefore {\tan }^4\theta =\frac{b^2}{a^2}\Rightarrow {\tan }^2\theta =\frac{b}{a}\Rightarrow \tan \theta =\pm \sqrt{\frac{b}{a}}$
For ${\tan }^2\theta =\frac{b}{a},$ Put in equation $\left(ii\right)$
$Z=a^2(1+\frac{b}{a})+b^2(1+\frac{a}{b})$
$\Rightarrow Z=a^2+2ab+b^2=(a+b{)}^2$
Put minimum value of $Z$ in equation $\left(i\right)$
$\therefore l_{max}=\frac{a^2-b^2}{a+b}=a-b$