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Mathematics · Ellipse

JEE Advanced 2022 — Paper 1 — Question 18

Consider the ellipse x24+y23=1\frac{x^{2}}{4}+\frac{y^{2}}{3}=1. Let H(α,0),0<α<2H(\alpha, 0), 0<\alpha<2, be a point. A straight line drawn through HH parallel to

yy-axis crosses the ellipse and its auxiliary circle at points EE and FF respectively, in the first quadrant.

The tangents 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\phi =\frac{\pi }{4}, then the area of the triangle FGHFGH is(P)(3−1)48\frac{{{(\sqrt{3}-1)}^{4}}}{8}
(II)If ϕ=π3\phi =\frac{\pi }{3}, then the area of the triangle FGHFGH is(Q)1
(III)If ϕ=π6\phi =\frac{\pi }{6}, then the area of the triangle FGHFGH is(R)34\frac{3}{4}
(IV)If ϕ=π12\phi =\frac{\pi }{12}, then the area of the triangle FGHFGH is(S)123\frac{1}{2\sqrt{3}}
(T)332\frac{3\sqrt{3}}{2}
  1. Option A:

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

  2. Option B:

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

  3. Option C:

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

    Correct
  4. Option D:

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

Answer: C

Step-by-step solution

Equation of auxiliary circle x2+y2=4x^{2}+y^{2}=4

∴\therefore Let F be (2cos⁡θ,2sin⁡θ)(2 \cos \theta, 2 \sin \theta)

∴E\therefore \mathrm{E} is (2cos⁡θ,3sin⁡θ)(2 \cos \theta, \sqrt{3} \sin \theta)

Equation of tangent at E,xcos⁡θ2+ysin⁡θ3=1\mathrm{E}, \frac{\mathrm{x} \cos \theta}{2}+\frac{\mathrm{y} \sin \theta}{\sqrt{3}}=1

It cuts x -axis at (2sec⁡θ,0)(2 \sec \theta, 0)

∴G\therefore \mathrm{G} is (2sec⁡θ,0)(2 \sec \theta, 0) H is (2cos⁡θ,0)(2 \cos \theta, 0) and F(2cos⁡θ,2sin⁡θ)\mathrm{F}(2 \cos \theta, 2 \sin \theta)

∴\therefore Area of Δ\Delta FGH is 12×2sin⁡θ(2sec⁡θ−2cos⁡θ)\frac{1}{2} \times 2 \sin \theta(2 \sec \theta-2 \cos \theta) =2sin⁡θ(sec⁡θ−cos⁡θ)=2 \sin \theta(\sec \theta-\cos \theta)

If θ=π4\theta=\frac{\pi}{4},

area =2×12(2−12)=1=2 \times \frac{1}{\sqrt{2}}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)=1

If θ=π3\theta=\frac{\pi}{3},

area =2×32(2−12)=332=2 \times \frac{\sqrt{3}}{2}\left(2-\frac{1}{2}\right)=\frac{3 \sqrt{3}}{2}

If θ=π6\theta=\frac{\pi}{6},

area =2×12(23−32)=123=2 \times \frac{1}{2}\left(\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}\right)=\frac{1}{2 \sqrt{3}}

If θ=π12\theta=\frac{\pi}{12},

area =2×3−122(223+1−3+122)=(3−18)4=2 \times \frac{\sqrt{3}-1}{2 \sqrt{2}}\left(\frac{2 \sqrt{2}}{\sqrt{3}+1}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)=\left(\frac{\sqrt{3}-1}{8}\right)^{4}.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse
Topic
Tangents & Normals to ellipse, chord of conatct