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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Ellipse

Let the length of the latus rectum of an ellipse with its major...

Let the length of the latus rectum of an ellipse with its major axis along xx -axis and centre at the origin, be 8 . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

Options

  1. A(42,22)(4 \sqrt{2}, 2 \sqrt{2})
  2. B(43,22)(4 \sqrt{3}, 2 \sqrt{2})
  3. C(43,23)(4 \sqrt{3}, 2 \sqrt{3})
  4. D(42,23)(4 \sqrt{2}, 2 \sqrt{3})

Correct answer

B. (43,22)(4 \sqrt{3}, 2 \sqrt{2})

Step-by-step solution

Let the ellipse be x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 Then, 2b2a=8,2ae−b2and\frac{2 b^{2}}{a}=8,2 a e-b^{2} andb^{2}=a^{3}\left(1-e^{2}\right)⇒a=8,b2=32\Rightarrow a=8, b^{2}=32 Then, the equation of the ellipse x264+y232=1\frac{x^{2}}{64}+\frac{y^{2}}{32}=1 Hence, the point (43,22)(4 \sqrt{3}, 2 \sqrt{2}) lies on the ellipse.
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