JEE Advanced · Mathematics · 14. Ellipse

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 $(P)$ $\frac{{(\sqrt{3} - 1)}^{4}}{8}$

$(II)$ 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 $(S)$ $\frac{1}{2 \sqrt{3}}$

$(T)$ $\frac{3 \sqrt{3}}{2}$

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 $(I) \to (Q); (II) \to (S); (III) \to (Q); (IV) \to (P)$

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Given $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ Let $α \equiv 2 \cos ϕ$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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	List- $I$		List- $II$
$(I)$	If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	$(P)$	$\frac{{(\sqrt{3} - 1)}^{4}}{8}$
$(II)$	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	$(S)$	$\frac{1}{2 \sqrt{3}}$
		$(T)$	$\frac{3 \sqrt{3}}{2}$