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 $\left(3\sqrt{2},2\right)$, is ________________.

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{th}\mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is}\frac{x^2}{a^2}-\frac{y^2}{a^2}=1 \\ \mathrm{which}\mathrm{passes}\mathrm{through}\left(3,0\right)\mathrm{and}\left(3\sqrt{2},2\right) \\ \mathrm{i}.\mathrm{e}.\frac{{\left(3\right)}^2}{a^2}-\frac{{\left(a\right)}^2}{b^2}=1 \\ \mathrm{i}.\mathrm{e}.a^2=9 \\ \mathrm{and}\frac{{\left(3\sqrt{2}\right)}^2}{a^2}-\frac{{\left(4\right)}^2}{b^2}=1 \\ \mathrm{i}.\mathrm{e}.\frac{9\times 2}{9}-\frac{4}{b^2}=1 \\ \mathrm{i}.\mathrm{e}.2-\frac{4}{b^2}=1 \\ \mathrm{i}.\mathrm{e}.b^2=4 \\ \therefore \mathrm{eccentricity}\mathrm{is}\sqrt{\frac{a^2+b^2}{a^2}} \\ =\sqrt{\frac{9+4}{9}} \\ \mathrm{i}.\mathrm{e}.\mathrm{eccentricity}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}. \end{gathered}$$