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Q.
The equation of the hyperbola whose vertices are $(\pm 2,0)$ and foci are $(\pm 3,0)$ is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 .$ Sum of $a^{2}$ and $b^{2}$ is
Conic Sections
Solution:
Since the foci are on $ x$-axis, the equation of the
hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Given: vertices are $(\pm 2,0), a=2$
Also, since foci are $(\pm 3,0), c=3$ and
$b^{2}=c^{2}-a^{2}=9-4=5$
$\therefore $ the equation of the hyperbola is : $\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$