The equation of the hyperbola with vertices at $(0, \pm 6)$ and eccentricity $\frac{5}{3}$ is ........... and its foci are ........... .
Let the equation of the hyperbola be $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Then vertices $=(0, \pm b)=(0, \pm 6)$
$$\begin{array}{llrl} \therefore & b =6 \text { and } e=5 / 3 \\ \because & e =\sqrt{1+\frac{a^2}{b^2}} \Rightarrow \frac{25}{9}=1+\frac{a^2}{36} \\ \Rightarrow & \frac{25-9}{9} =\frac{a^2}{36} \Rightarrow 16=\frac{a^2}{4} \Rightarrow a^2=48 \end{array}$$
So, the equation of hyperbola is,
$$\begin{aligned} \frac{-x^2}{48}+\frac{y^2}{36} & =1 \Rightarrow \frac{y^2}{36}-\frac{x^2}{48}=1 \\ \text { Foci } & =(0, \pm \text { be })=\left(=0, \pm \frac{5}{3} \times 6\right)=(0, \pm 10) \end{aligned}$$
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