JEE Challenger
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MathHardAdv 2022 • Paper 1

Area of Triangle Formed by Ellipse Tangent and Auxiliary Circle

Consider the ellipse

x24+y23=1.\frac{x^2}{4} + \frac{y^2}{3} = 1.

Let H(α,0)H(\alpha, 0), 0<α<20 < \alpha < 2, be a point. A straight line drawn through HH parallel to the yy-axis crosses the ellipse and its auxiliary circle at points EE and FF respectively, in the first quadrant. The tangent to the ellipse at the point EE intersects the positive xx-axis at a point GG. Suppose the straight line joining FF and the origin makes an angle ϕ\phi with the positive xx-axis.

List-IList-II(I) If ϕ=π4, then the area of the triangle FGH is(P) (3−1)48(II) If ϕ=π3, then the area of the triangle FGH is(Q) 1(III) If ϕ=π6, then the area of the triangle FGH is(R) 34(IV) If ϕ=π12, then the area of the triangle FGH is(S) 123(T) 332\begin{array}{ll} \text{List-I} & \text{List-II} \\[4pt] \text{(I) If } \phi = \frac{\pi}{4}\text{, then the area of the triangle } FGH \text{ is} & \text{(P) } \frac{(\sqrt{3}-1)^4}{8} \\[4pt] \text{(II) If } \phi = \frac{\pi}{3}\text{, then the area of the triangle } FGH \text{ is} & \text{(Q) } 1 \\[4pt] \text{(III) If } \phi = \frac{\pi}{6}\text{, then the area of the triangle } FGH \text{ is} & \text{(R) } \frac{3}{4} \\[4pt] \text{(IV) If } \phi = \frac{\pi}{12}\text{, then the area of the triangle } FGH \text{ is} & \text{(S) } \frac{1}{2\sqrt{3}} \\[4pt] & \text{(T) } \frac{3\sqrt{3}}{2} \end{array}

The correct option is:

Options

A

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

B

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

C

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

Correct
D

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

Topics & Concepts

Conic SectionsEllipse

Step-by-Step Solution

To determine the correct matching option, we analyze the given geometric configuration step-by-step.

1. Parametric Representation of Points

The equation of the ellipse is given by:

x24+y23=1\frac{x^2}{4} + \frac{y^2}{3} = 1

Here, a=2a = 2 and b=3b = \sqrt{3}. The auxiliary circle of the ellipse is:

x2+y2=a2  ⟹  x2+y2=4x^2 + y^2 = a^2 \implies x^2 + y^2 = 4

Let the eccentric angle be ϕ∈(0,π2)\phi \in \left(0, \frac{\pi}{2}\right).

  • Since the point FF lies on the auxiliary circle in the first quadrant and the line OFOF makes an angle ϕ\phi with the positive xx-axis, the coordinates of FF are:
    F=(2cos⁡ϕ,2sin⁡ϕ)F = (2\cos\phi, 2\sin\phi)

  • The line through H(α,0)H(\alpha, 0) parallel to the yy-axis is x=αx = \alpha. Since this line passes through FF, we have:
    α=2cos⁡ϕ  ⟹  H=(2cos⁡ϕ,0)\alpha = 2\cos\phi \implies H = (2\cos\phi, 0)

  • The point EE lies on the ellipse in the first quadrant with the same xx-coordinate α=2cos⁡ϕ\alpha = 2\cos\phi, so its coordinates are:
    E=(2cos⁡ϕ,3sin⁡ϕ)E = (2\cos\phi, \sqrt{3}\sin\phi)

2. Tangent to the Ellipse and Point GG

The equation of the tangent to the ellipse x24+y23=1\frac{x^2}{4} + \frac{y^2}{3} = 1 at E(2cos⁡ϕ,3sin⁡ϕ)E(2\cos\phi, \sqrt{3}\sin\phi) is:
x(2cos⁡ϕ)4+y(3sin⁡ϕ)3=1  ⟹  xcos⁡ϕ2+ysin⁡ϕ3=1\frac{x(2\cos\phi)}{4} + \frac{y(\sqrt{3}\sin\phi)}{3} = 1 \implies \frac{x\cos\phi}{2} + \frac{y\sin\phi}{\sqrt{3}} = 1

Point GG is the intersection of this tangent with the positive xx-axis. Setting y=0y = 0:
xcos⁡ϕ2=1  ⟹  x=2sec⁡ϕ\frac{x\cos\phi}{2} = 1 \implies x = 2\sec\phi
Thus, the coordinates of GG are: G=(2sec⁡ϕ,0)G = (2\sec\phi, 0)

3. Area of Triangle FGHFGH

The vertices of △FGH\triangle FGH are: F=(2cos⁡ϕ,2sin⁡ϕ),G=(2sec⁡ϕ,0),H=(2cos⁡ϕ,0)F = (2\cos\phi, 2\sin\phi), \quad G = (2\sec\phi, 0), \quad H = (2\cos\phi, 0)

Notice that HH and GG lie on the xx-axis, and the line FHFH is perpendicular to the xx-axis (since both FF and HH have xx-coordinate 2cos⁡ϕ2\cos\phi). Therefore, △FGH\triangle FGH is a right-angled triangle at HH.

  • Base length (HGHG):
    HG=2sec⁡ϕ−2cos⁡ϕ=2(1−cos⁡2ϕcos⁡ϕ)=2sin⁡2ϕcos⁡ϕHG = 2\sec\phi - 2\cos\phi = 2\left(\frac{1 - \cos^2\phi}{\cos\phi}\right) = \frac{2\sin^2\phi}{\cos\phi}

  • Height (FHFH):
    FH=2sin⁡ϕFH = 2\sin\phi

The area Δ\Delta of △FGH\triangle FGH is:

Δ=12×HG×FH=12(2sin⁡2ϕcos⁡ϕ)(2sin⁡ϕ)=2sin⁡3ϕcos⁡ϕ=2sin⁡2ϕtan⁡ϕ\Delta = \frac{1}{2} \times HG \times FH = \frac{1}{2} \left(\frac{2\sin^2\phi}{\cos\phi}\right) (2\sin\phi) = \frac{2\sin^3\phi}{\cos\phi} = 2\sin^2\phi \tan\phi

4. Evaluating the Cases

  • Case (I): ϕ=π4\phi = \frac{\pi}{4}
    Δ=2sin⁡2(π4)tan⁡(π4)=2(12)2(1)=1\Delta = 2 \sin^2\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) = 2 \left(\frac{1}{\sqrt{2}}\right)^2 (1) = 1
    So, (I) →\rightarrow (Q).

  • Case (II): ϕ=π3\phi = \frac{\pi}{3}
    Δ=2sin⁡2(π3)tan⁡(π3)=2(32)2(3)=2(34)3=332\Delta = 2 \sin^2\left(\frac{\pi}{3}\right) \tan\left(\frac{\pi}{3}\right) = 2 \left(\frac{\sqrt{3}}{2}\right)^2 (\sqrt{3}) = 2 \left(\frac{3}{4}\right)\sqrt{3} = \frac{3\sqrt{3}}{2}
    So, (II) →\rightarrow (T).

  • Case (III): ϕ=π6\phi = \frac{\pi}{6}
    Δ=2sin⁡2(π6)tan⁡(π6)=2(12)2(13)=2(14)(13)=123\Delta = 2 \sin^2\left(\frac{\pi}{6}\right) \tan\left(\frac{\pi}{6}\right) = 2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{\sqrt{3}}\right) = 2 \left(\frac{1}{4}\right)\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{2\sqrt{3}}
    So, (III) →\rightarrow (S).

  • Case (IV): ϕ=π12\phi = \frac{\pi}{12}
    Using sin⁡(π12)=3−122\sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{3}-1}{2\sqrt{2}} and tan⁡(π12)=2−3=(3−1)22\tan\left(\frac{\pi}{12}\right) = 2-\sqrt{3} = \frac{(\sqrt{3}-1)^2}{2}:
    Δ=2(3−122)2((3−1)22)=2((3−1)28)((3−1)22)=(3−1)48\Delta = 2 \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 \left(\frac{(\sqrt{3}-1)^2}{2}\right) = 2 \left(\frac{(\sqrt{3}-1)^2}{8}\right) \left(\frac{(\sqrt{3}-1)^2}{2}\right) = \frac{(\sqrt{3}-1)^4}{8}
    So, (IV) →\rightarrow (P).

Conclusion

The correct mapping is:

(I)→(Q);(II)→(T);(III)→(S);(IV)→(P)\text{(I)} \rightarrow \text{(Q)}; \quad \text{(II)} \rightarrow \text{(T)}; \quad \text{(III)} \rightarrow \text{(S)}; \quad \text{(IV)} \rightarrow \text{(P)}

This corresponds to Option C.

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