Question

If the foci of ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide, then $b^2$ =

• A. 4
• B. 5
• C. 8
• D. 9

Answer 1

The correct option is B 5

$\frac{x^2}{25}+\frac{y^2}{16}=1$

let the focii of ellipse be $(\pm a_1{e}_1,o)$

$a_1=5,e_1=\sqrt{1-\frac{16}{25}}$

$e_1=\frac{3}{5}$

focii of hyperbola $(\pm b_2{e}_2,o)$

$a_1{e}_1=b_2{e}_2$

$3=b_2\times \sqrt{1+\frac{4}{b_2^2}}$

$\frac{9}{b^2}=1+\frac{4}{b^2}$

$\Rightarrow \frac{5}{b^2}=1$

$b^2=5$