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 $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
$\Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}\Rightarrow 9+b^2=13$
$\Rightarrow b=\pm 2$
Now, from $e{q}^n\left(1\right),$ we have
$\frac{K^2}{9}-\frac{4}{4}=1\left(\because b=\pm 2\right)$
$\Rightarrow K^2=18$