Q58 of 62 Page 11

Foci (0, ± 13), the conjugate is of length 24.


Since the foci are on y-axis, the equation of the hyperbola is of the form

= 1

Given: foci are (0, ± 13), c = 13

Length of conjugate axis = 2b = 24 ⇒ b = 12

or b2 = c2 – a2 or 144 ⇒ 169 – a2

or a2 = 25 or a = 5

The equation of the hyperbola is

= 1

or 144y2 - 25x2 = 3600.

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