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

Open MARKS Web Download App

Search any question & find its solution

Question: Answered & Verified by Expert

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

MathematicsHyperbolaJEE Main

Options:

A $\frac{256}{9}$
B $\frac{16}{3}$
C $9$
D $3$

Solution:

2795 Upvotes Verified Answer

The correct answer is:

\frac{256}{9}

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

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.

Use on iOS or Desktop Download from Google Play