Question

The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which

​​​​​​​passes through the points $(3,0)$ and $(3\sqrt{2},2)$ is $\underset{–}{}$

Answer 1

Given equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\because$ Hyperbola passes through points $(3,0)$ and $(3\sqrt{2},2)$

$\therefore$ These points will satisfy given equation of hyperbola.

$\therefore \frac{3^2}{a^2}-\frac{0^2}{b^2}=1$

$\Rightarrow \frac{9}{a^2}=1$

$\Rightarrow a^2=9\dots \left(1\right)$

And $\frac{(3\sqrt{2}{)}^2}{a^2}-\frac{2^2}{b^2}=1$

$\Rightarrow \frac{18}{9}-\frac{4}{b^2}=1$ (from (1) )

$\Rightarrow 2-\frac{4}{b^2}=1$

$\Rightarrow \frac{4}{b^2}=1$

$\Rightarrow b^2=4\dots \left(2\right)$

$\because \text{Eccentricity},e=\sqrt{1+\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1+\frac{4}{9}}$ (from (1) and (2))

$\Rightarrow e=\frac{\sqrt{13}}{3}$