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Q.
If the equation of hyperbola is $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$, then
Conic Sections
Solution:
The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.
Thus, $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ has transverse axis along $x$ -axis of length $6$ .