Foci of hyperbola 9{x^2} - 16{y^2} = 144 are

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10-2. Parabola, Ellipse, Hyperbola

easy

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

A

$( \pm 4,\;0)$

B

$(0,\; \pm 4)$

C

$( \pm 5,\;0)$

D

$(0,\; \pm 5)$

Solution

(c) The equation of hyperbola is $\frac{{{x^2}}}{{16}} – \frac{{{y^2}}}{9} = 1$

Now ${b^2} = {a^2}({e^2} – 1)$

$ \Rightarrow \,e = \frac{5}{4}$

Hence foci are $( \pm \,ae,\,0)$

==> $\left( { \pm \,4.\frac{5}{4},\,0} \right)$

$i.e.$, $( \pm \,5,\,0)$.

Standard 11

Mathematics

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