Question

The greatest distance of a normal to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ from its centre is:

• A. $a-b$
• B. $a+b$
• C. $\frac{a-b}{2}$
• D. $\frac{a+b}{2}$

Answer 1

The correct option is A $a-b$

Equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $P(acos\theta ,bsin\theta )$ be a point on the ellipse.
Equation of normal to ellipse at P is
$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=a^2-b^2$
Let S be the distance of normal from centre(0,0) .
$S=|\frac{-(a^2-b^2)}{\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }}|$
S is maximum when $\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }$ is minimum.
Minimum value of $a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta =(a+b{)}^2$
Hence maximum distance $=\frac{a^2-b^2}{a+b}=a-b$