Question

From a variable point of an ellipse $\frac{x^2}{d^2}+\frac{y^2}{b^2}=1$ normal is drawn to the ellipse. Find the maximum distance of the normal from the centre of the ellipse.

• A. $a+b.$
• B. $a-b.$
• C. $a^2-b^2.$
• D. $-a+b.$

Answer 1

The correct option is B $a-b.$

Given equation of ellipse is
$\frac{x^2}{d^2}+\frac{y^2}{b^2}=1$
Equation of normal to ellipse at $(a\cos t,b\sin t)$ is
$ax\sec t-bycosect=a^2-b^2$
If $p$ be the length of perpendicular from (0,0), then
$p=\frac{a^2-b^2}{\sqrt{(a^{2}se{c}^{2}t+b^{2}cose{c}^{2}t)}}$
Now p will be maximum if its denominator $\sqrt{a^2{\sec }^{2}t+b^{2}cose{c}^{2}t}$ is minimum as its numerator is constant.
Let $z=\sqrt{a^{2}se{c}^{2}t+b^{2}cose{c}^{2}t}$
$\frac{dz}{dt}=\frac{2a^2{\sec }^{2}t\tan t-2b^{2}cose{c}^{2}t\cot t}{2\sqrt{a^2{\sec }^{2}t+b^{2}cose{c}^{2}t}}$
For maxima or minima,
$\frac{dz}{dt}=0$
$\Rightarrow a^2\frac{\sin t}{{\cos }^{3}t}=\frac{b^2\cos t}{{\sin }^{3}t}$
$\Rightarrow {\tan }^{4}t=\frac{b^2}{a^2}$
$\Rightarrow {\tan }^{2}t=\frac{b}{a}$
Also, $\frac{d^z}{d{t}^2}>0$ for ${\tan }^{2}t=\frac{b}{a}$
So, minimum value of $z$ is
$Minz=\sqrt{a^2}(1+\frac{b}{a})+b^2(1+\frac{a}{b})$
$\Rightarrow Minz=a+b$
$\therefore$ Maximum value of $p=(a^2-b^2)/(a+b)=a-b.$