Given:

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

Here, the foci are on y-axis.

The equation of the hyperbola is of the form \( \dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1\)

Since, the foci are (0, ±13), so, c = 13

Since, the length of the conjugate axis is 24,

2b = 24

b = \( \dfrac{24}{2}\)

= 12

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 12^{2} = 13^{2}

a^{2} = 169 – 144

= 25

The equation of the hyperbola is \(\dfrac{ y^2}{25} –\dfrac{ x^2}{144} = 1\)

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