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. | List-I | List-II | | --- | --- | | (I) If $\phi = \frac{\pi}{4}$, then the area of the triangle $FGH$ is | (P) $\frac{(\sqrt{3} - 1)^{4}}{8}$ | | (II) If $\phi = \frac{\pi}{3}$, then the area of the triangle $FGH$ is | (Q) $1$ | | (III) If $\phi = \frac{\pi}{6}$, then the area of the triangle $FGH$ is | (R) $\frac{3}{4}$ | | (IV) If $\phi = \frac{\pi}{12}$, then the area of the triangle $FGH$ is | (S) $\frac{1}{2\sqrt{3}}$ | | | (T) $\frac{3\sqrt{3}}{2}$ | The correct option is:

Consider the ellipse x²/4 + y²/3 = 1. Let H(α, 0), 0 < α < 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 φ with the positive x-axis. | List-I | List-II | | --- | --- | | (I) If φ = π/ $\text{Given: ellipse } \dfrac{x^2}{4} + \dfrac{y^2}{3} = 1,\ \text{auxiliary circle } x^2 + y^2 = 4$ $H(\alpha,0),\ 0 < \alpha < 2;\ E \text{ on ellipse, } F \text{ on aux.\ circle (first quadrant)}$ $\text{tangent at } E \text{ meets } x\text{-axis at } G;\quad \angle FOx = \phi$ $\text{To find: match each } \phi \text{ with the area of } \triangle FGH$ $1.\ \text{get } F \text{ on aux.\ circle, so } H \text{ and } E$ $2.\ \text{tangent at } E \Rightarrow G$ $3.\ \text{area of } \triangle FGH \text{ vs } \phi,\ \text{then match}$ $\text{ellipse}: a = 2,\ b = \sqrt{3};\quad \text{aux.\ circle}: x^2 + y^2 = 4$ $F = (2\cos\phi,\ 2\sin\phi)\ \Rightarrow\ \alpha = 2\cos\phi$ $H = (2\cos\phi,\ 0)$ $E = (2\cos\phi,\ \sqrt{3}\sin\phi)$ $\dfrac{x\,x_1}{4} + \dfrac{y\,y_1}{3} = 1 \ \Rightarrow\ \dfrac{x\cos\phi}{2} + \dfrac{y\sin\phi}{\sqrt{3}} = 1$ $y=0:\ \dfrac{x\cos\phi}{2}=1 \Rightarrow G=(2\sec\phi,\ 0)$ $GH = 2\sec\phi - 2\cos\phi = \dfrac{2\sin^2\phi}{\cos\phi}$ $[\triangle FGH] = \dfrac12 \cdot \dfrac{2\sin^2\phi}{\cos\phi}\cdot 2\sin\phi = \dfrac{2\sin^3\phi}{\cos\phi}$ $= 2\tan\phi\,\sin^2\phi$ $\phi=\dfrac{\pi}{4}:\ \tan\phi=1,\ \sin^2\phi=\dfrac12$ $[\triangle FGH] = 2\cdot 1\cdot \dfrac12 = 1 \ \Rightarrow\ \text{(I)} \to \text{(Q)}$ $\phi=\dfrac{\pi}{3}:\ \tan\phi=\sqrt{3},\ \sin^2\phi=\dfrac34$ $[\triangle FGH] = 2\cdot\sqrt{3}\cdot\dfrac34 = \dfrac{3\sqrt{3}}{2} \ \Rightarrow\ \text{(II)} \to \text{(T)}$ $\phi=\dfrac{\pi}{6}:\ \tan\phi=\dfrac{1}{\sqrt{3}},\ \sin^2\phi=\dfrac14$ $[\triangle FGH] = 2\cdot\dfrac{1}{\sqrt3}\cdot\dfrac14 = \dfrac{1}{2\sqrt{3}} \ \Rightarrow\ \text{(III)} \to \text{(S)}$ $\phi=\dfrac{\pi}{12}:\ \tan\phi=2-\sqrt{3},\ \sin^2\phi=\dfrac{2-\sqrt{3}}{4}$ $[\triangle FGH] = 2(2-\sqrt{3})\cdot\dfrac{2-\sqrt{3}}{4} = \dfrac{(2-\sqrt{3})^2}{2}$ $= \dfrac{(\sqrt{3}-1)^4}{8} \ \Rightarrow\ \text{(IV)} \to \text{(P)}$ $\text{(I)}\to\text{(Q)},\ \text{(II)}\to\text{(T)},\ \text{(III)}\to\text{(S)},\ \text{(IV)}\to\text{(P)}$ $\boxed{\text{Answer: option (3)}}$