Question

If the eccentricity of a hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{b^{2}}=1,$ which passes through $\left(K, 2\right),$ is $\frac{\sqrt{13}}{3},$ then the value of $K^2$ is

• A. $18$
• B. $8$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

Given hyperbola is
$\frac{x^{2}}{9}-\frac{y^{2}}{b^{2}}=1$
Since this passes through $\left(K, 2\right),$ therefore
$\frac{K^{2}}{9}-\frac{4}{b^{2}}=1\,...\left(1\right)$
Also, given $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{\sqrt{13}}{3}$
$\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{\sqrt{13}}{3} \Rightarrow 9+b^{2}=13$
$\Rightarrow b=\pm2$
Now, from $eq^{n} \left(1\right),$ we have
$\frac{K^{2}}{9}-\frac{4}{4}=1\,\left(\because b=\pm2\right)$
$\Rightarrow K^{2}=18$