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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse

JEE Advanced 2012 · Paper 1 · Mathematics · Ellipse

The ellipse E_ 1 : x^ 2 9 + y^ 2 4 =1 is inscribed in a rectangle R...

The ellipse E1:x29+y24=1E_{1}: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is inscribed in a rectangle RR whose sides are parallel to the coordinate axes. Another ellipse E2E_{2} passing through the point (0,4)(0,4) circumscribes the rectangle RR. The eccentricity of the ellipse E2E_{2} isThe ellipse E_ 1 : x^ 2 9 + y^ 2 4 =1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes.

Options

  1. A22\frac{\sqrt{2}}{2}
  2. B32\frac{\sqrt{3}}{2}
  3. C12\frac{1}{2}
  4. D34\frac{3}{4}

Correct answer

C. 12\frac{1}{2}

Step-by-step solution

As rectangle ABCDA B C D circumscribed the ellipse x29+y24=1,∴A=(3,2)\frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \therefore A=(3,2) The ellipse E_ 1 : x^ 2 9 + y^ 2 4 =1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes.
Let the ellipse circumscribing the rectangle ABCDA B C D is x2a2+y2b2=1...(i)\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1...(i) Given that ellipse (i) passes through (a,4)(a, 4) ∴b2=16\therefore b^{2}=16 Also ellipse (i) passes through A(3,2)A(3,2) ∴9a2+416=1⇒a2=12\therefore \quad \frac{9}{a^{2}}+\frac{4}{16}=1 \Rightarrow a^{2}=12 ∴e=1−1216=14=12\therefore \quad e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}
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