Q82 / 225 · JEE Main 2003 · Mathematics

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is

  • A
    9
  • B
    1
  • C
    5
  • Correct answer
    D
    7

Solution

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

Hence, the correct answer is option: 'D'.

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