If the eccentricity of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$, which passes through $(K, 2)$, is…
If the eccentricity of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$, which passes through $(K, 2)$, is $\frac{\sqrt{13}}{3}$, then the value of $K^2$ is
18
8
1
2
Solution
Given hyperbola is
$
\frac{x^2}{9}-\frac{y^2}{b^2}=1
$
Since this passes through $(K, 2)$, therefore
$
\frac{K^2}{9}-\frac{4}{b^2}=1
$
Also, given $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
$
\begin{aligned}
& \Rightarrow \sqrt{1+\frac{b^2}{9}}=\frac{\sqrt{13}}{3} \Rightarrow 9+b^2=13 \\
& \Rightarrow b=\pm 2
\end{aligned}
$
Now, from $\mathrm{eq}^{\mathrm{n}}$ (1), we have
$
\begin{aligned}
& \frac{K^2}{9}-\frac{4}{4}=1 \quad(\because b=\pm 2) \\
& \Rightarrow \quad K^2=18
\end{aligned}
$