If $e_1$ and $e_2$ are respectively the eccentricities of the curves $9x^2 - 16y^2 - 144 = 0$ and $9x^2 - 16y^2 + 144 = 0$,then find the value of $\frac{e_1^2 e_2^2}{e_1^2 + e_2^2}$.

  • A
    $\sqrt{2}$
  • B
    $1$
  • C
    $\sqrt{3}$
  • D
    $2$

Explore More

Similar Questions

The equation of the normal to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at $(-4, 0)$ is

If two points $P$ and $Q$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with centre $C$ are such that $CP$ is perpendicular to $CQ$,where $a < b$,then the value of $\frac{1}{(CP)^2} + \frac{1}{(CQ)^2}$ is:

If the centre,vertex,and focus of a hyperbola are $(0, 0)$,$(4, 0)$,and $(6, 0)$ respectively,then the equation of the hyperbola is

The values of $m$ for which the line $y=mx+2$ is a tangent to the hyperbola $4x^2-9y^2=36$ are

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k=$

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