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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Ellipse

The length of the minor axis (along y -axis) of an ellipse in the...

The length of the minor axis (along yy-axis) of an ellipse in the standard form is 43\dfrac{4}{\sqrt{3}}. If this ellipse touches the line x+6y=8x + 6y = 8 then its eccentricity is:

Options

  1. A12113\dfrac{1}{2}\sqrt{\dfrac{11}{3}}
  2. B56\sqrt{\dfrac{5}{6}}
  3. C1253\dfrac{1}{2}\sqrt{\dfrac{5}{3}}
  4. D13113\dfrac{1}{3}\sqrt{\dfrac{11}{3}}

Correct answer

A. 12113\dfrac{1}{2}\sqrt{\dfrac{11}{3}}

Step-by-step solution

Let equation of ellipse be x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1.

Here, 2b=43⇒b=232b = \dfrac{4}{\sqrt{3}} ⇒ b = \dfrac{2}{\sqrt{3}}

Equation of tangents ≡y=mx±a2m2+b2≡ y = mx \pm \sqrt{{a}^{2}{m}^{2} + {b}^{2}}

Comparing with ≡y=−x6+43≡ y = - \dfrac{x}{6} + \dfrac{4}{3}

m=−16m = - \dfrac{1}{6} and a2m2+b2=169{a}^{2}{m}^{2} + {b}^{2} = \dfrac{16}{9}

⇒a236+43=169⇒ \dfrac{{a}^{2}}{36} + \dfrac{4}{3} = \dfrac{16}{9}

⇒a236=169−43=49⇒ \dfrac{{a}^{2}}{36} = \dfrac{16}{9} - \dfrac{4}{3} = \dfrac{4}{9}

⇒a2=16⇒ {a}^{2} = 16

Now, e=1−b2a2e = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}}

⇒e=1−43×16=1112⇒ e = \sqrt{1 - \dfrac{4}{3 \times 16}} = \sqrt{\dfrac{11}{12}}

⇒e=12113⇒ e = \dfrac{1}{2}\sqrt{\dfrac{11}{3}}

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