Question

Normal is drawn to the ellipse $\frac{x^2}{27}+y^2=1$ at a Point ($3\sqrt{3}cos\theta ,sin\theta$) where $0<\theta <\frac{\pi }{2}$. The value of $\theta$ such that the area of triangle formed by normal and coordinate axes is maximum, is

Answer 1

$(3\sqrt{3}\cos \theta ,\sin \theta )$

$x=3\sqrt{3}\cos \theta ,y=\sin \theta$

$x=a\cos \theta ,y=b\sin \theta \Rightarrow a=3\sqrt{3},b=1$

Equation of normal, is given by:

$\frac{b^2(y-y_1)}{y_1}=\frac{a^2(x-x_1)}{x_1}$

$\frac{1(y-\sin \theta )}{\sin \theta }=\frac{27(x-3\sqrt{3}\cos \theta )}{3\sqrt{3}\cos \theta }$

$\frac{1(y-\sin \theta )}{\sin \theta }=\frac{3\sqrt{3}(x-3\sqrt{3}\cos \theta )}{\cos \theta }$

$y\cos \theta -\sin \theta \cos \theta =\left(3\sqrt{3}\sin \theta \right)x-27\sin \theta \cos \theta$

$\left(3\sqrt{3}\sin \theta \right)x-\left(\cos \theta \right)y-26\sin \theta \cos \theta =0$

is the required normal equation.

$26\sin \theta \cos \theta \ne 0$

$\therefore 3\sqrt{3}\sin \theta -\cos \theta =0$

$\frac{\sin \theta }{\cos \theta }=\frac{1}{3\sqrt{3}}$

$\therefore \tan \theta =\frac{1}{3\sqrt{3}}$