Q59 / 90 · JEE Main 2012 (19 May Online) · Mathematics

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ coincide with the foci of the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$, then $b^2$ is equal to

  • A
    8
  • B
    10
  • Correct answer
    C
    7
  • D
    9

Solution

Given equation of ellipse is $$ \frac{x^2}{16}+\frac{y^2}{b^2}=1 $$ $$ \begin{aligned} & \text { eccentricity }=e=\sqrt{1-\frac{b^2}{16}} \\ & \text { foci: } \pm a e=\pm 4 \sqrt{1-\frac{b^2}{16}} \end{aligned} $$ Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ $$ \begin{aligned} & \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \\ & \text { eccentricity }=e=\sqrt{1+\frac{81}{25} \times \frac{25}{144}}=\sqrt{1+\frac{81}{144}} \\ & =\sqrt{\frac{225}{144}}=\frac{15}{12} \\ & \text { foci: } \pm a e=\pm \frac{12}{5} \times \frac{15}{12}=\pm 3 \\ & \end{aligned} $$ Since, foci of ellipse and hyperbola coincide & \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \\ $$

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

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