The foci of hyperbola $4 x^2-9 y^2-1=0$ are
The foci of hyperbola $4 x^2-9 y^2-1=0$ are
- $( \pm \sqrt{13}, 0)$
- $\left( \pm \frac{\sqrt{13}}{6}, 0\right)$
- $\left(0, \pm \frac{\sqrt{3}}{6}\right)$
- None of these
Solution
Given, Hyperbola $=4 x^2-9 y^2-1=0$
$\begin{aligned}
& \Rightarrow \frac{x^2}{\left(\frac{1}{4}\right)}-\frac{y^2}{\left(\frac{1}{9}\right)}=1 \\
& \Rightarrow \frac{x^2}{\frac{1}{2}^2}-\frac{y^2}{\frac{1}{3}^2}=1
\end{aligned}$
We know that. if hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. then Eccentricity:
$e=\sqrt{1+\frac{b^2}{a^2}}$
$\begin{aligned}
& \text {and focus } \equiv( \pm \text{be}, 0)=\left( \pm \frac{1}{2} \times \frac{\sqrt{1 / 3}}{3}, 0\right)=\left( \pm \frac{\sqrt{13}}{6}, 0\right)
\end{aligned}$
Asked in: BITSAT 2024 (Memory Based Paper 3)
Practice more Hyperbola questions on Aicharya