Consider the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ Let $H(\alpha, 0), 0

Question

Consider the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ Let $H(\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis. The correct option is :

Professor PC Thomas Classes · Answer

(c) : $E(2 \cos \phi, \sqrt{3} \sin \phi)$\ The equation of $E G$ :\ $\frac{x}{2} \cos \phi+\frac{y}{\sqrt{3}} \sin \phi=1$\ Hence $G \equiv\left(\frac{2}{\cos \phi}, 0ight)$ and $H \equiv(2 \cos \phi, 0)$\ $\operatorname{ar}(	riangle F G H)=\frac{1}{2} H G 	imes F H$\ $=\frac{1}{2} \cdot\left(\frac{2}{\cos \phi}-2 \cos \phiight) \cdot 2 \sin \phi=\frac{2 \sin ^{3} \phi}{\cos \phi}$\ So, $f(\phi)=\frac{2 \sin ^{3} \phi}{\cos \phi}$\ Now, $f\left(\frac{\pi}{4}ight)=2 \cdot \frac{\left(\frac{1}{\sqrt{2}}ight)^{3}}{\frac{1}{\sqrt{2}}}=1$,\ $f\left(\frac{\pi}{3}ight)=2 \cdot \frac{\left(\frac{\sqrt{3}}{2}ight)^{3}}{\left(\frac{1}{2}ight)}=\frac{3 \sqrt{3}}{2}$,\ $f\left(\frac{\pi}{6}ight)=2 \cdot \frac{\left(\frac{1}{2}ight)^{3}}{\frac{\sqrt{3}}{2}}=\frac{1}{2 \sqrt{3}}$\ and $f\left(\frac{\pi}{12}ight)=2 \cdot \frac{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}ight)^{3}}{\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}ight)}=\frac{(\sqrt{3}-1)^{4}}{8}$\