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, a2 + b2 = c2
a2 + 122 = 132
a2 = 169 – 144
= 25
The equation of the hyperbola is \(\dfrac{ y^2}{25} –\dfrac{ x^2}{144} = 1\)
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