MCQOPTIONS
Saved Bookmarks
MCQOPTIONS
| 1. |
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» D. 9 | |