Foci of hyperbola frac{{{x^2}}}{{16}} - frac{{{{(y - 2)}^2}}}{9} = 1 are

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10-2. Parabola, Ellipse, Hyperbola

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Foci of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{{(y - 2)}^2}}}{9} = 1$ are

A

$(5, 2) (-5, 2)$

B

$(5, 2) (5, -2)$

C

$(5, 2) (-5, -2)$

D

None of these

Solution

(a) $a = 4,\,b = 3$

$⇒$ $\frac{9}{{16}} = ({e^2} – 1)$

$⇒$ $e = \frac{5}{4}$

Vertex is $(0, 2)$.

Hence focus is $( \pm ae,\,2) = ( \pm 5,\,2)$.

Standard 11

Mathematics

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