9.Vertices (0, ± 3), foci (0, ± 5)

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Solution

The given coordinates of vertices are ( 0,±3 ) and coordinates of focus are ( 0,±5 ) .

Since the vertices are on the y axis, the equation of the hyperbola is represented as,

y 2 a 2 − x 2 b 2 =1 , where y is the transverse axis.(1)

Since y axis is the transverse axis, coordinates of Foci = (0,±c)

∴c=5

Since y axis is the transverse axis, coordinates of vertices = (0,±a)

∴a=3

a 2 + b 2 = c 2 3 2 + b 2 = 5 2 b 2 =25−9 b 2 =16 b=4

Substitute the values of a and b in equation (1)

y 2 9 − x 2 16 =1

Thus, the equation of hyperbola with vertices (0,±3) and foci ( 0,±5 ) is y 2 9 − x 2 16 =1 .

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

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

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

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

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

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

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

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

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

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

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

(0, \pm 3)

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Find the equation of the hyperbola satisfying the given condition,

Vertices $(0, \pm 3), f o c i (0, \pm 5)$

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