If the eccentricity of the hyperbola (x^2)/16-(y^2)/(b^2)=-1 is 5/4, then b^2 is equal to

Correct answer: (A) 256/9. y^2b^2-x^216=1 For the hyperbola, e=√(1+16b^2) ⇒√(1+16b^2)=5/4⇒b^2=256/9

Q17 / 40 · NTA Abhyas (JEE Main) 2020 · Mathematics · Hyperbola

If the eccentricity of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = - 1$ is $\frac{5}{4},$ then $b^{2}$ is equal to

$\frac{y^{2}}{b^{2}} - \frac{x^{2}}{16} = 1$

For the hyperbola, $e = \sqrt{1 + \frac{16}{b^{2}}}$
$\Rightarrow \sqrt{1 + \frac{16}{b^{2}}} = \frac{5}{4} \Rightarrow b^{2} = \frac{256}{9}$

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