If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$ respectively,then $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $3$

Explore More

Similar Questions

$A$ tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{2} = 1$ meets the $x-$axis at $P$ and the $y-$axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on

For different values of $\alpha$,the locus of the point of intersection of the two straight lines $\sqrt{3} x - y - 4 \sqrt{3} \alpha = 0$ and $\sqrt{3} \alpha x + \alpha y - 4 \sqrt{3} = 0$ is

Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $8 \sqrt{\frac{5}{3}}$. If for $H$,the length of the latus rectum is $l$ and the product of the focal distances of the point $P$ is $m$,then $9l^2 + 6m$ is equal to :-

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 $\frac{e_1^2 e_2^2}{e_1^2 + e_2^2}$.

The line $21x + 5y = k$ touches the hyperbola $7x^2 - 5y^2 = 232$. 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