If the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ are $\alpha$ and $\beta$,then $\frac{\cot(\alpha/2)}{\tan(\beta/2)}=$

  • A
    $\frac{4}{3}$
  • B
    $-9$
  • C
    $9$
  • D
    $\frac{4}{5}$

Explore More

Similar Questions

If $P(\theta)$ and $Q\left(\frac{\pi}{2}+\theta\right)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the locus of the midpoint of $PQ$ is $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$,then $\frac{a+b}{\alpha+\beta}=$

The length of the chord of the ellipse $\frac{x^2}{4} + y^2 = 1$ formed on the line $y = x + 1$ is

Let $S \equiv \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ and $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}-1=0$ be two intersecting ellipses. If $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ are their points of intersection,then $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$

If $x \cos \alpha + y \sin \alpha = 4$ is a tangent to $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$,then the value of $\alpha$ is

Consider the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The portion of the tangent at any point of the curve intercepted between the point of contact and the directrix subtends at the corresponding focus an angle of

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