JEE Advanced 2022 Paper 1 Q18 Mathematics Coordinate Geometry Ellipse Hard

JEE Advanced 2022 Paper 1 · Q18 · Ellipse

Consider the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1$. 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.

List-I: (I) If $\phi=\pi/4$, then the area of the triangle $FGH$ is (II) If $\phi=\pi/3$, then the area of the triangle $FGH$ is (III) If $\phi=\pi/6$, then the area of the triangle $FGH$ is (IV) If $\phi=\pi/12$, then the area of the triangle $FGH$ is

List-II: (P) $(\sqrt{3}-1)^{4}/8$ (Q) $1$ (R) $3/4$ (S) $1/(2\sqrt{3})$ (T) $3\sqrt{3}/2$

The correct option is:

  1. A. (I)$\to$(R); (II)$\to$(S); (III)$\to$(Q); (IV)$\to$(P)
  2. B. (I)$\to$(R); (II)$\to$(T); (III)$\to$(S); (IV)$\to$(P)
  3. C. (I)$\to$(Q); (II)$\to$(T); (III)$\to$(S); (IV)$\to$(P)
  4. D. (I)$\to$(Q); (II)$\to$(S); (III)$\to$(Q); (IV)$\to$(P)
Reveal answer + step-by-step solution

Correct answer:C

Solution

Auxiliary circle: $x^{2}+y^{2}=4$. With parameter $\phi$ on auxiliary circle, $F=(2\cos\phi,2\sin\phi)$, $H=(2\cos\phi,0)$, $E=(2\cos\phi,\sqrt{3}\sin\phi)$ on ellipse. Tangent at $E$ on ellipse: $\dfrac{x\cos\phi}{2}+\dfrac{y\sin\phi}{\sqrt{3}}=1$. At $y=0$: $x=2/\cos\phi$, so $G=(2/\cos\phi,0)$. Then $HG=2/\cos\phi-2\cos\phi=2\sin^{2}\phi/\cos\phi$. $FH=2\sin\phi$. Area $=\dfrac{1}{2}\cdot HG\cdot FH=\dfrac{1}{2}\cdot\dfrac{2\sin^{2}\phi}{\cos\phi}\cdot 2\sin\phi=\dfrac{2\sin^{3}\phi}{\cos\phi}$. (I) $\phi=\pi/4$: area $=\dfrac{2(1/\sqrt{2})^{3}}{1/\sqrt{2}}=\dfrac{2/(2\sqrt{2})}{1/\sqrt{2}}=\dfrac{1/\sqrt{2}}{1/\sqrt{2}}=1$ (Q). (II) $\phi=\pi/3$: area $=\dfrac{2(\sqrt{3}/2)^{3}}{1/2}=\dfrac{2\cdot 3\sqrt{3}/8}{1/2}=\dfrac{3\sqrt{3}/4}{1/2}=3\sqrt{3}/2$ (T). (III) $\phi=\pi/6$: area $=\dfrac{2(1/2)^{3}}{\sqrt{3}/2}=\dfrac{1/4}{\sqrt{3}/2}=\dfrac{1}{2\sqrt{3}}$ (S). (IV) $\phi=\pi/12$: $\sin(\pi/12)=(\sqrt{6}-\sqrt{2})/4$, $\cos(\pi/12)=(\sqrt{6}+\sqrt{2})/4$. After simplification area $=(\sqrt{3}-1)^{4}/8$ (P). Match: (I)Q,(II)T,(III)S,(IV)P = option (C).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →