The equation of the hyperbola with vertices (3, 0), (–3, 0) and semi-latus rectum 4, is given by

4 x^{2} - 3 y^{2} + 36 = 0

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4 x^{2} - 3 y^{2} + 12 = 0

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4 x^{2} - 3 y^{2} - 36 = 0

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Solution

The correct option is C $4 x^{2} - 3 y^{2} - 36 = 0$
We have, a = 3 and $\frac{b^{2}}{a} = 4$ $\Rightarrow$ $b^{2} = 12$
Hence, the equation of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{12} = 1$ $\Rightarrow$ $4 x^{2} - 3 y^{2} = 36$

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The vertices of a hyperbola are (2, 0), (–2, 0) and the foci are (3, 0), (–3, 0). The equation of the hyperbola is

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Find the equation of the hyperbola with foci (0, $\pm$ 3) and vertices $(0, \pm \frac{\sqrt{11}}{2})$ .

In each of the following find the equation of the hyperbola satisfying the given conditions.:

$(i) v e r t i c e s (\pm 2, 0), f o c i (\pm 3, 0)$

$(i i) v e r t i c e s (0, \pm, 5), f o c i (0, \pm, 8)$

$(i i i) v e r t i c e s (0, \pm, 3), f o c i (0, \pm, 5)$

$(i v) v e r t i c e s (\pm 5, 0), t r a n s v e r s e a x i s = 8$

$(v) f o c i (0, \pm 13), c o n j u g a t e a x i s = 24$

$(v i) f o c i (\pm 3 \sqrt{5}), t h e l a t u s - r e c t u m = 8$

$(v i i) f o c i (\pm, 4, 0), t h e l a t u s - r e c t u m = 12$

$(v i i i) v e r t i c e s (0, \pm 6), e = \frac{5}{3} .$

$(i x) f o c i (0, \pm \sqrt{1} 0), p a s s i n g t h r o u g h (2, 3)$

$(x) f o c i (0, \pm 12), l a t u s - r e c t u m = 36$

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