Question

The maximum distance of the normal to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ from its centre is:

• A. $\frac{1}{2}$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is C $1$

Ellipse : $\frac{x^2}{9}+\frac{y^2}{4}=1$

Equation of the normal,

$\frac{{ax}^{}}{\cos \theta }-\frac{{by}^{}}{\sin \theta }=a^2-b^2\therefore \frac{{3x}^{}}{\cos \theta }-\frac{{2y}^{}}{\sin \theta }=5$

Or, ${3x}^{}\sin \theta -{2y}^{}\cos \theta =5\cos \theta \sin \theta$

Distance from origin d= $\frac{|0+0-5\cos \theta \sin \theta |}{\sqrt{9{\left(\cos \theta \right)}^2+{4\left(\sin \theta \right)}^2}}$

Or, d=$\frac{5}{\sqrt{9{\left(\csc \theta \right)}^2+4{\left(\sec \theta \right)}^2}}$

To maximize d we need to minimize the denominator.

$E=9{\left(\csc \theta \right)}^2+4{\left(\sec \theta \right)}^{2}then,\frac{dE}{d\theta }=-18{\left(\csc \theta \right)}^2\cot \theta +8{\left(\sec \theta \right)}^2\tan \theta ForMinimizing,\frac{dE}{d\theta }=0\therefore -18{\left(\csc \theta \right)}^2\cot \theta +8{\left(\sec \theta \right)}^2\tan \theta =0Or,18{\left(\csc \theta \right)}^2\cot \theta =8{\left(\sec \theta \right)}^2\tan \theta Or,{\left(\tan \theta \right)}^4=\frac{9}{4}Or,\tan \theta =\sqrt{\frac{3}{2}}\therefore \csc \theta =\sqrt{\frac{5}{3}}and\sec \theta =\sqrt{\frac{5}{2}}$

On putting the values in d we get,

$d=\frac{5}{\sqrt{15+10}}Or,d=1$