Question

If 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 b² is

• A. 1
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

7

Answer 2

Given equation of ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$.

Here a² = 16 ⇒ a = 4.

Foci of the ellipse are (± ae, 0) i.e., (± 4e, 0).

Given equation of hyperbola is

$\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}or\frac{x^{2}}{144} - \frac{y^{2}}{81} = 1$.

Here a² = $\frac{144}{25} \Rightarrow a = \frac{12}{5}andb^{2} = \frac{81}{25} \Rightarrow b \Rightarrow \frac{9}{5}$.

Also b² = a² $(e_{1}^{2} - 1) \Rightarrow \frac{81}{25} = \frac{144}{25}(e_{1}^{2} - 1)$

$\Rightarrow e_{1}^{2} - 1 = \frac{81x25}{25x144} = \frac{9}{16}$.

∴ $e_{1}^{2} = \frac{25}{16}i.e.e_{1} = \frac{5}{4}$.

Foci of hyperbola = $\left( \pm \frac{12}{5}x\frac{5}{4},0 \right) = ( \pm 3,0)$

According to the question, we have

4e = 3 ⇒ e = ¾.

For ellipse, b² = a+2 ^{(1 - e}2^).

⇒ b² = 16 $\left( 1 - \frac{9}{16} \right)$ = 16 x $\frac{7}{16}$= 7.