JEE Advanced 2022 · Paper 1 · Mathematics Q18

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MathematicsList match+3 / -1Question 18 of 54

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:

List-I

I
If $\phi = \frac\pi4$ , then the area of the triangle $FGH$ is
II
If $\phi = \frac\pi3$ , then the area of the triangle $FGH$ is
III
If $\phi = \frac\pi6$ , then the area of the triangle $FGH$ is
IV
If $\phi = \frac\pi{12}$ , then the area of the triangle $FGH$ is

List-II

P
$\frac{(\sqrt3-1)^4}{8}$
Q
$1$
R
$\frac34$
S
$\frac1{2\sqrt3}$
T
$\frac{3\sqrt3}{2}$

(I) → (R); (II) → (S); (III) → (Q); (IV) → (P)

(I) → (R); (II) → (T); (III) → (S); (IV) → (P)

(I) → (Q); (II) → (T); (III) → (S); (IV) → (P)

(I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

Answer and worked solution

Official answer: C

The points. Here $a=2$ , $b=\sqrt3$ . Use the eccentric angle $\phi$ :

$F=(2\cos\phi,2\sin\phi)$ and $E=(2\cos\phi,\sqrt3\sin\phi)$ ;
$H=(2\cos\phi,0)$ .

The tangent at E is $\frac{x\cos\phi}2+\frac{y\sin\phi}{\sqrt3}=1$ , so $G=\left(\frac2{\cos\phi},0\right)$ .

Area. Triangle FGH has a right angle at H:
$\text{Area}=\frac12\cdot\left(\frac2{\cos\phi}-2\cos\phi\right)\cdot2\sin\phi=\frac{2\sin^3\phi}{\cos\phi}.$

Values.

$\phi=\frac\pi4$ : area 1 → Q.
$\phi=\frac\pi3$ : $\frac{3\sqrt3}2$ → T.
$\phi=\frac\pi6$ : $\frac1{2\sqrt3}$ → S.
$\phi=\frac\pi{12}$ : $\frac{2\sin^3\phi}{\cos\phi}=8\sin^4 15^\circ=\frac{(2-\sqrt3)^2}2=\frac{(\sqrt3-1)^4}8$ → P.

Topics:Ellipse Tangents and normals to conics Trigonometric ratios and identities

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JEE Advanced 2022 · Paper 1, Mathematics question 18