Question

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

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

Answer 1

The correct option is B $a-b$

Equation of any normal to the ellipse is:

$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=ae$

Distance from the center is: $d=\frac{ae}{\sqrt{\frac{a^2}{{\left(\cos \theta \right)}^2}+\frac{b^2}{{\left(\sin \theta \right)}^2}}}$ -------(1)

$a,e,b$ are fixed for a given ellipse. So, to maximize $d$ we need to minimize the denominator

$\therefore E=a^2({\sec \theta )}^2+b^2{\left(co\sec \theta \right)}^2$

$\frac{DE}{D\theta }=2a^2({\sec \theta )}^3\tan \theta -{2b}^2{\left(co\sec \theta \right)}^3\cot \theta =0$

$\Rightarrow 2a^2({\sec \theta )}^3\tan \theta ={2b}^2{\left(co\sec \theta \right)}^3\cot \theta$

$\Rightarrow {\left(\tan \theta \right)}^4=\frac{b^2}{a^2}$

$\Rightarrow \tan \theta =\sqrt{\frac{b}{a}}$

$\therefore \sin \theta =\sqrt{\frac{b}{a+b}}$ and

$\cos \theta =\sqrt{\frac{a}{a+b}}$

Putting these values in equation 1 we get:

$d=\frac{ae}{\sqrt{\frac{a^2(a+b)}{a}+\frac{b^2(a+b)}{b}}}$

$\Rightarrow d=\frac{ae}{(a+b)}$

$\Rightarrow d=\frac{ae(a-b)}{(a+b)(a-b)}$

$\Rightarrow d=a-b$