The foci of the hyperbola (9x^{2} - 16y^{2} = 144) are | MATH - HYPERBOLA | SolveeIt

The foci of the hyperbola $9x^{2} - 16y^{2} = 144$ are

$( \pm 4,0)$

$(0, \pm 4)$

$( \pm 5,0)$

$(0, \pm 5)$

Answer

$( \pm 5,0)$

Explanation

The equation of hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ .

Now, $b^{2} = a^{2}(e^{2} - 1)$ ⇒ $9 = 16(e^{2} - 1)$ ⇒ $e = \frac{5}{4}$ . Hence foci

are $( \pm ae,0)$ = $\left( \pm 4.\frac{5}{4},0 \right)$ i.e., $( \pm 5,0)$

Question: The foci of the hyperbola \(9x^{2} - 16y^{2} = 144\) are...