Question

The maximum distance of the centre of the ellipse $\frac{x^2}{81}+\frac{y^2}{25}=1$ from a normal to the ellipse is _____________.

Answer 1

$\frac{x^2}{81}+\frac{y^2}{25}=1$

centre$\equiv (0,0)$

Equation of tangent in parametric form

$\frac{x9\cos \theta }{81}+\frac{y5\sin \theta }{25}=1$

$\frac{x\cos \theta }{9}+\frac{y\sin \theta }{5}=1$

$\Rightarrow y=\frac{-5x\cos \theta }{9\sin \theta }+\frac{5}{\sin \theta }$

Slope of normal$\times$Slope of Tangent$=-1$

$m_n=\frac{9\sin \theta }{5\cos \theta }$Equation of normal; $y-5\sin \theta =\frac{9\sin \theta }{5\cos \theta }(x-9\cos \theta )$$5y\cos \theta -25\sin \theta \cos \theta =9x\sin \theta -81\sin \theta \cos \theta$

$5ycosec\theta -25=9x\sec \theta -81$

$9x\sec \theta -5ycosec\theta -56$

Distance form centre$=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|=\left|\frac{0+0+-56}{\sqrt{(9\sec \theta {)}^2+(5cosec\theta {)}^2}}\right|$

$=\frac{56}{\sqrt{81{\sec }^2\theta +25cose{c}^2\theta }}-------\left(A\right)$

To find the maximum value

$81{\sec }^2\theta +25cose{c}^2\theta =m$

$\frac{dm}{d\theta }=0$

$81\times 2{\sec }^2\theta \tan \theta -25\times 2cose{c}^2\theta \cot \theta =0$

$81\times 2{\sec }^2\theta \tan \theta =25\times 2cose{c}^2\theta \cot \theta$

$81\times 2{\sec }^2\theta \frac{\sec \theta }{cosec\theta }=25\times 2cose{c}^2\theta \frac{cosec\theta }{\sec \theta }$

${\tan }^4\theta =\frac{25}{81}$

${\tan }^2\theta =\frac{5}{9}$

$\therefore {\cot }^2\theta =\frac{9}{5}$

Substituting in A

$=\frac{56}{\sqrt{81(1+{\tan }^2\theta )+25(1+{\cot }^2\theta )}}$

$=\frac{56}{\sqrt{81\left(\frac{14}{9}\right)+25\left(\frac{14}{5}\right)}}$

$=\frac{56}{\sqrt{126+70}}=\frac{56}{\sqrt{196}}=\frac{56}{14}=4$