Question

Normal at $(2\cos \theta ,\sin \theta )$ on the ellipse $x^2+4y^2=4$ intersects it again at $(2\cos \varphi ,\sin \varphi )$ if $\sin \varphi =$

• A. $-\cos \theta \left(\frac{7-9{\sin }^2\theta }{1+15{\sin }^2\theta }\right)$
• B. $-\sin \theta \left(\frac{7+9{\sin }^2\theta }{1+15{\sin }^2\theta }\right)$
• C. $-\sin \theta \left(\frac{7+11{\sin }^2\theta }{1-13{\sin }^2\theta }\right)$
• D. $-\cos \theta \left(\frac{3+5{\sin }^2\theta }{1+7{\sin }^2\theta }\right)$

Answer 1

Answer: (B)

Equation of normal at $P(2\cos \theta ,\sin \theta )$ is
$\frac{2x}{\cos \theta }-\frac{y}{\sin \theta }=3$
or $x=\frac{\cos \theta }{2}\left(3+\frac{y}{\sin \theta }\right)$
Squaring and putting the value of $x^2$ from the equation of ellipse,
$\frac{{\cos }^2\theta }{4}{\left(3+\frac{y}{\sin \theta }\right)}^2+4y^2-4=0$
Above equation is quadratic in $y$ which has two roots $y_1$ and $y_2$, such that
$y_1{y}_2=1^2\cdot \sin \theta \sin \varphi =\frac{\frac{9}{4}{\cos }^2\theta -4}{\frac{{\cos }^2\theta }{4{\sin }^2\theta }+4}$
$={\sin }^2\theta \frac{9{\cos }^2\theta -16}{{\cos }^2\theta +16{\sin }^2\theta }$
$=-{\sin }^2\theta \left(\frac{7+9{\sin }^2\theta }{1+15{\sin }^2\theta }\right)$
$\Rightarrow \sin \varphi =-\sin \theta \left(\frac{7+9{\sin }^2\theta }{1+15{\sin }^2\theta }\right)$