If $e$ and $e'$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,then $\frac{1}{e^2} + \frac{1}{e'^2} = \dots$

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    None of these

Explore More

Similar Questions

The tangent to the hyperbola $xy = c^2$ at the point $P(ct, c/t)$ intersects the $x$-axis at $T$ and the $y$-axis at $T'$. The normal to the hyperbola at $P$ intersects the $x$-axis at $N$ and the $y$-axis at $N'$. If the areas of the triangles $PNT$ and $PN'T'$ are $\Delta$ and $\Delta'$ respectively,then $\frac{1}{\Delta} + \frac{1}{\Delta'}$ is:

$A$ 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

What kind of hyperbola does the equation $9x^2 - 16y^2 - 18x + 32y - 151 = 0$ represent?

If $P(\theta)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $S$ and $S^{\prime}$ are the foci of the hyperbola,then $SP \cdot S^{\prime}P =$

If the product of the slopes of the tangents drawn from an external point $P(h, k)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a constant $k^2$,then the locus of $P$ 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