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Q.
Normal at $(2 \cos \theta, \sin \theta)$ on the ellipse $x^{2}+4 y^{2}=4$ intersects it again at $(2 \cos \phi, \sin \phi)$ if $\sin \phi=$
Conic Sections
Solution:
Equation of normal at $P(2 \cos \theta, \sin \theta)$ is
$\frac{2 x}{\cos \theta}-\frac{y}{\sin \theta}=3$
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,
$\frac{\cos ^{2} \theta}{4}\left(3+\frac{y}{\sin \theta}\right)^{2}+4 y^{2}-4=0$
Above equation is quadratic in $y$ which has two roots $y_{1}$ and $y_{2}$, such that
$y_{1} y_{2}=1^{2} \cdot \sin \theta \sin \phi=\frac{\frac{9}{4} \cos ^{2} \theta-4}{\frac{\cos ^{2} \theta}{4 \sin ^{2} \theta}+4}$
$=\sin ^{2} \theta \frac{9 \cos ^{2} \theta-16}{\cos ^{2} \theta+16 \sin ^{2} \theta} $
$=-\sin ^{2} \theta\left(\frac{7+9 \sin ^{2} \theta}{1+15 \sin ^{2} \theta}\right)$
$\Rightarrow \sin \phi =-\sin \theta\left(\frac{7+9 \sin ^{2} \theta}{1+15 \sin ^{2} \theta}\right)$