JEE Advanced 2022 — Paper 1 — MATH-III solution

JEE Advanced2022Paper 1MATH-III

Consider the ellipse

$\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1.$

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 $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	(P) $\frac{( 3 - 1 ) ^{4}}{8}$
(II) If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is	(Q) $1$
(III) If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is	(R) $\frac{3}{4}$
(IV) If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is	(S) $\frac{1}{2 3}$
	(T) $\frac{3 3}{2}$

The correct option is:

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

Solution

The ellipse has semi-axes $a = 2$ , $b = 3$ . Its auxiliary circle is $x^{2} + y^{2} = 4$ . With $H = (α, 0)$ , the vertical line through $H$ meets the auxiliary circle at $F = (α, 4 - α^{2})$ in the first quadrant. Writing $α = 2 cos ϕ$ (where $ϕ$ is the eccentric angle), $F = (2 cos ϕ, 2 sin ϕ)$ .

On the ellipse the corresponding point is $E = (2 cos ϕ, 3 sin ϕ)$ .

Tangent at $E$ to the ellipse: $\frac{x cos ϕ}{2} + \frac{y sin ϕ}{3} = 1$ . Setting $y = 0$ gives $G = (2 sec ϕ, 0)$ .

Triangle $F G H$ is right-angled at $H$ , with horizontal leg $H G = 2 sec ϕ - 2 cos ϕ = \frac{2 ( 1 - cos ^{2} ϕ )}{cos ϕ} = \frac{2 sin ^{2} ϕ}{cos ϕ}$ , vertical leg $H F = 2 sin ϕ$ .

$Δ = \frac{1}{2} \cdot H G \cdot H F = \frac{1}{2} \cdot \frac{2 sin ^{2} ϕ}{cos ϕ} \cdot 2 sin ϕ = \frac{2 sin ^{3} ϕ}{cos ϕ} = 2 sin^{2} ϕ tan ϕ = (1 - cos 2 ϕ) tan ϕ .$

(I) $ϕ = \frac{π}{4}$ : $Δ = 2 \cdot \frac{1}{2} \cdot 1 = 1$ . $\to$ (Q).

(II) $ϕ = \frac{π}{3}$ : $Δ = 2 \cdot \frac{3}{4} \cdot 3 = \frac{3 3}{2}$ . $\to$ (T).

(III) $ϕ = \frac{π}{6}$ : $Δ = 2 \cdot \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2 3}$ . $\to$ (S).

(IV) $ϕ = \frac{π}{12}$ : Using $sin \frac{π}{12} = \frac{6 - 2}{4}$ and $cos \frac{π}{12} = \frac{6 + 2}{4}$ , $tan \frac{π}{12} = 2 - 3$ . Compute $2 sin^{2} \frac{π}{12} tan \frac{π}{12} = (1 - cos \frac{π}{6}) (2 - 3) = (1 - \frac{3}{2}) (2 - 3) = \frac{( 2 - 3 ) ^{2}}{2} = \frac{( 3 - 1 ) ^{4}}{8}$ (using the identity $(2 - 3)^{2} = \frac{( 3 - 1 ) ^{4}}{4}$ which follows from $(3 - 1)^{2} = 4 - 23 = 2 (2 - 3)$ ). $\to$ (P).

Matching: (I) $\to$ (Q); (II) $\to$ (T); (III) $\to$ (S); (IV) $\to$ (P), which is option (C).

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