Consider the ellipse x ^ 2 4 + y ^ 2 3 = 1 . Let H(α, 0), 0 List- I List- II (I) If ϕ = (π)/(4) , then the area of the triangle FGH is (P) ( 3 - 1) ^ 4 8 (II) If ϕ = (π)/(3) , then the area of the triangle FGH is (Q) 1

Correct answer: C. (I) → (Q); (II) → (T); (III) → (S); (IV) → (P) Given x ^ 2 4 + y ^ 2 3 = 1 Let α ≡ 2cosϕ Tangent at E(2cosϕ, 3 sinϕ) to the ellipse is (xcosϕ)/(2) + ysinϕ 3 = 1 This intersect x -axis at G(2secϕ, 0) Drawing the figure as per the information, we get, Area of triangle FGH = (1)/(2)(2secϕ - 2cosϕ)2sinϕ Δ = 2 ^ 2 ϕ . tanϕ Δ = (1 - 2ϕ) · tanϕ I. If ϕ = (π)/(4), Δ = 1 → (Q) II. If ϕ = (π)/(3), Δ = 2 · ( 3 2 ) ^ 2 · 3 = 3 3 2 → (T) III. If ϕ = (π)/(6), Δ = 2 · ((1)/(2)) ^ 2 · 1 3 = 1 2 3 → (S) IV. If ϕ = (π)/(12), Δ = (1 - 3 2 ) · (2 - 3 ) = (2 - 3 ) ^ 2 2 = ( 3 - 1) ^ 4 8 → (P)

JEE Advanced 2022Paper 1

Exam: JEE Advanced
Year: 2022
Paper: Paper 1
Subject: Mathematics
Chapter: Ellipse

JEE Advanced 2022 · Paper 1 · Mathematics · Ellipse

Consider the ellipse x ^ 2 4 + y ^ 2 3 = 1 . Let H(α, 0), 0 List- I...

Consider the ellipse $\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1$ . Let $H\left(\alpha, 0\right), 0 < \alpha < 2$ , 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 $F and the origin makes an angle$ ϕ $with the positive$ x $-axis.</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td> </td> <td>List-$ I $</td> <td> </td> <td>List-$ II $</td> </tr> <tr> <td>\left(I\right)$ If $ϕ = \dfrac{\pi}{4}$ , then the area of the triangle $FGH$ is \left(P\right) $</td> <td>$ \dfrac{{\left(\sqrt{3} - 1\right)}^{4}}{8} $</td> </tr> <tr> <td>\left(II\right)$ If $ϕ = \dfrac{\pi}{3}$ , then the area of the triangle $FGH$ is \left(Q\right) $</td> <td>$ 1 $</td> </tr> <tr> <td>\left(III\right)$ If $ϕ = \dfrac{\pi}{6}$ , then the area of the triangle $FGH$ is \left(R\right) $</td> <td>$ \dfrac{3}{4} $</td> </tr> <tr> <td>\left(IV\right)$ If $ϕ = \dfrac{\pi}{12}$ , then the area of the triangle $FGH$ is \left(S\right) $</td> <td>$ \dfrac{1}{2\sqrt{3}} $</td> </tr> <tr> <td> </td> <td> </td> <td>\left(T\right)$ $\dfrac{3\sqrt{3}}{2}$

The correct option is:

Consider the ellipse x ^ 2 4 + y ^ 2 3 = 1 . Let H(α, 0), 0 List- I List- II (I) If ϕ = (π)/(4) , then the area of the

Options

A $\left(I\right) \to \left(R\right); \left(II\right) \to \left(S\right); \left(III\right) \to \left(Q\right); \left(IV\right) \to \left(P\right)$
B $\left(I\right) \to \left(R\right); \left(II\right) \to \left(T\right); \left(III\right) \to \left(S\right); \left(IV\right) \to \left(P\right)$
C $\left(I\right) \to \left(Q\right); \left(II\right) \to \left(T\right); \left(III\right) \to \left(S\right); \left(IV\right) \to \left(P\right)$
D $\left(I\right) \to \left(Q\right); \left(II\right) \to \left(S\right); \left(III\right) \to \left(Q\right); \left(IV\right) \to \left(P\right)$

Correct answer

C. $\left(I\right) \to \left(Q\right); \left(II\right) \to \left(T\right); \left(III\right) \to \left(S\right); \left(IV\right) \to \left(P\right)$

Step-by-step solution

Given $\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1$
Let $\alpha ≡ 2cosϕ$
Tangent at $ $E\left(2cosϕ, \sqrt{3}sinϕ\right)$ to the ellipse is $\dfrac{xcosϕ}{2} + \dfrac{ysinϕ}{\sqrt{3}} = 1$
This intersect $x$ -axis at $G\left(2secϕ, 0\right)$
Drawing the figure as per the information, we get,
Area of triangle $FGH = \dfrac{1}{2}\left(2secϕ - 2cosϕ\right)2sinϕ$
$\Delta = 2\sin^{2}ϕ . tanϕ$
$\Delta = \left(1 - \cos 2ϕ\right) \cdot tanϕ$
I. If $ϕ = \dfrac{\pi}{4}, \Delta = 1 \to \left(Q\right)$
II. If $ϕ = \dfrac{\pi}{3}, \Delta = 2 \cdot {\left(\dfrac{\sqrt{3}}{2}\right)}^{2} \cdot \sqrt{3} = \dfrac{3\sqrt{3}}{2} \to \left(T\right)$
III. If $ϕ = \dfrac{\pi}{6}, \Delta = 2 \cdot {\left(\dfrac{1}{2}\right)}^{2} \cdot \dfrac{1}{\sqrt{3}} = \dfrac{1}{2\sqrt{3}} \to \left(S\right)$
IV. If $ϕ = \dfrac{\pi}{12}, \Delta = \left(1 - \dfrac{\sqrt{3}}{2}\right) \cdot \left(2 - \sqrt{3}\right) = \dfrac{{\left(2 - \sqrt{3}\right)}^{2}}{2} = \dfrac{{\left(\sqrt{3} - 1\right)}^{4}}{8} \to \left(P\right)

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