Question

If the foci of the ellipse 16x2 + 7y2 = 112 and of the hyperbola $\frac{y^{2}}{144}–\frac{x^{2}}{a^{2}} = \frac{1}{25}$coincide, then a =

• A. 3
• B. 9
• C. 81
• D. 8

Answer 1

Answer: (B)

9

Answer 2

For the ellipse$\frac{x^{2}}{7} + \frac{y^{2}}{16} = 1$ , e2 = $\frac{a^{2}–b^{2}}{a^{2}}$ = $\frac{9}{16}$

\ e =$\frac{3}{4}$

CS = ae $\frac{3}{4}$= 4 ×$\frac{3}{4}$= 3

For the hyperbola $\frac{y^{2}}{\frac{144}{25}}–\frac{x^{2}}{\frac{a^{2}}{25}}$= 1,

e¢2 = $\frac{a^{2} + b^{2}}{a^{2}} = \frac{\frac{144}{25} + \frac{a^{2}}{25}}{\frac{144}{25}}$

\ e¢ = $\frac{\sqrt{144 + a^{2}}}{12}$ and CS = ae¢

= $\frac{12}{5}$×$\frac{\sqrt{144 + a^{2}}}{12}$

CS being the same in both the cases$\frac{\sqrt{144 + a^{2}}}{5}$= 3

\ a2 = 81, a = 9