Find the equation of the hyperbola with eccentricity $e = 3/2$ and foci at $(\pm 2, 0)$.

  • A
    $\frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}$
  • B
    $\frac{x^2}{9} - \frac{y^2}{9} = \frac{4}{9}$
  • C
    $\frac{x^2}{4} - \frac{y^2}{9} = 1$
  • D
    None of these

Explore More

Similar Questions

If the eccentricities of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ are $e$ and $e_1$ respectively,then $\frac{1}{e^2} + \frac{1}{e_1^2} = $

If $\sqrt{5} y - \sqrt{8} = 0$ is the equation of the directrix of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$ and $\frac{\sqrt{5}}{2}$ is its eccentricity,then $\frac{1}{a} =$

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the reciprocal of the eccentricity of the ellipse $x^2+4y^2=4$. If the hyperbola passes through a focus of the ellipse,then:

If the distance between the foci and the distance between the directrices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are in the ratio $3: 2$,then $a: b$ is

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo