If $P(\frac{\pi}{6})$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,$S$ and $S^{\prime}$ are its foci,and $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,then the eccentricity $e$ is:

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

Explore More

Similar Questions

Let the points $P_1\left(\frac{\pi}{4}\right), P_2\left(\frac{3 \pi}{4}\right), P_3\left(\frac{5 \pi}{4}\right)$ and $P_4\left(\frac{7 \pi}{4}\right)$ given in parametric form,lie on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$. Then these four points in that order form

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $(x_i, y_i)$,for $i=1, 2, 3, 4$,then $y_1+y_2+y_3+y_4$ equals

If the eccentricity of a hyperbola is $\sqrt{3}$,then the eccentricity of its conjugate hyperbola 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 :-

Let tangents drawn from point $C(0,-b)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ touch the hyperbola at points $A$ and $B$. If $\Delta ABC$ is a right-angled triangle,then $\frac{a^2}{b^2}$ is equal to -

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