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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Mathematics
Chapter
Ellipse

JEE Main 2003 · Mathematics · Ellipse

The foci of the ellipse (x^2)/(16)+(y^2)/(b^2)=1 and the hyperbola...

The foci of the ellipse x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=1 and the hyperbola x2144−y281=125\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} coincide. Then the value of b2b^2 is

Options

  1. A9
  2. B1
  3. C5
  4. D7

Correct answer

D. 7

Step-by-step solution

\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \mathrm{a}=\sqrt{\frac{144}{25}}, \mathrm{~b}=\sqrt{\frac{81}{25}}, \mathrm{e}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}FociFoci=(3,0),focusofellipse, focus of ellipse=(3,0) \Rightarrow \mathrm{e}=\frac{3}{4} b^2=16\left(1-\frac{9}{16}\right)=7
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