Foci (±5, 0) and the transverse axis is of length 8.
Here, the foci are on x-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 (±5, 0), so, c = 5
Since, the length of the transverse axis is 8,
2a = 8
a = \( \dfrac{8}{2}\)
= 4
It is known that, a2 + b2 = c2
42 + b2 = 52
b2 = 25 – 16
= 9
The equation of the hyperbola is
\( \dfrac{x^2}{16}- \dfrac{y^2}{9}=1\)
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