The sum of the focal distances of any point on the conic $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is

  • A
    $10$
  • B
    $9$
  • C
    $41$
  • D
    $18$

Explore More

Similar Questions

An ellipse is drawn by taking a diameter of the circle $(x - 1)^2 + y^2 = 1$ as its semi-minor axis and a diameter of the circle $x^2 + (y - 2)^2 = 4$ as its semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes,then the equation of the ellipse is:

The angle between the tangents drawn from a point $(-3, 2)$ to the ellipse $4x^2 + 9y^2 - 36 = 0$ is

If the straight line $x \cos \alpha + y \sin \alpha = p$ touches the curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,then prove that $a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha = p^{2}$.

Difficult
View Solution

If any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ cuts off intercepts of length $h$ and $k$ on the axes,then $\frac{a^2}{h^2} + \frac{b^2}{k^2} = $

Difficult
View Solution

The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse $9x^2 + 4y^2 = 72$ at the point $(2, 3)$ with the $X$-axis 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