Question

The eccentricity of the hyperbola in the standard form $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$, passing through $\left(3,0\right)$ and $\left(3\sqrt{2},2\right)$ is

• A. $\frac{13}{3}$
• B. $\sqrt{13}$
• C. $\sqrt{3}$
• D. $\frac{\sqrt{13}}{3}$
• E. $\frac{5}{3}$

Answer 1

The correct option is D

$\frac{\sqrt{13}}{3}$

Explanation for the correct answer:

Finding the eccentricity:

The standard form of a hyperbola is $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$.

The hyperbola passes through $\left(3,0\right)$.

So, put $x=3$ and $y=0$ in $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$

$$\begin{gathered} \Rightarrow \left(\frac{3^2}{a^2}\right)+0=1 \\ \Rightarrow a^2=9 \end{gathered}$$

The hyperbola passes through $\left(3\sqrt{2},2\right)$.

So, put $x=3\sqrt{2}$ , $y=2$ and $a^2=9$ in $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$

$$\begin{gathered} \Rightarrow \left(\frac{{\left(3\sqrt{2}\right)}^2}{\left(9\right)}\right)-\left(\frac{{\left(2\right)}^2}{b^2}\right)=1 \\ \Rightarrow \frac{18}{9}-\frac{4}{b^2}=1 \\ \Rightarrow 2-\frac{4}{b^2}=1 \\ \Rightarrow \frac{4}{b^2}=1 \\ \Rightarrow b^2=4 \end{gathered}$$

The eccentricity of the hyperbola is given by $e=\sqrt{1+\frac{b^2}{a^2}}$.

Put $a^2=9,b^2=4$.

$$\begin{gathered} \Rightarrow e=\sqrt{1+\frac{4}{9}} \\ =\sqrt{\frac{13}{9}} \\ =\frac{\sqrt{13}}{3} \end{gathered}$$

Hence, the correct answer is option (D)