The equations of the directrices of the ellipse $16x^2 + 25y^2 = 400$ are

  • A
    $2x = \pm 25$
  • B
    $5x = \pm 9$
  • C
    $3x = \pm 10$
  • D
    None of these

Explore More

Similar Questions

If $4x+y+p=0$ $(p>0)$ is a tangent to the ellipse $x^2+3y^2=3$ and $16x+qy+14=0$ $(q>0)$ is a normal to the ellipse $x^2+8y^2=33$,then $p+q=$

The equation of the ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$ is

The area of the region enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is . . . . . . square units. (in $\pi$)

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 20y - 16 = 0$ is:

Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e = \frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle,whose vertex $P$ is on the circle $C$ and the side $QR$ of length $2$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ 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