In the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the equation of the diameter conjugate to the diameter $y = \frac{b}{a}x$ is:

  • A
    $y = - \frac{b}{a}x$
  • B
    $y = - \frac{a}{b}x$
  • C
    $x = - \frac{b}{a}y$
  • D
    None of these

Explore More

Similar Questions

Let $A_1, A_2, A_3$ be regions in the $XY$-plane defined by:
$A_1 = \{(x, y) : x^2 + 2y^2 \leq 1\}$
$A_2 = \{(x, y) : |x|^3 + 2\sqrt{2}|y|^3 \leq 1\}$
$A_3 = \{(x, y) : \max(|x|, \sqrt{2}|y|) \leq 1\}$
Then,

For the ellipse $25x^2 + 9y^2 - 150x - 90y + 225 = 0$,the eccentricity $e$ is: (in $/5$)

Let $F$ and $F'$ be the foci of the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ $(b < 2)$ and $B$ is one end of the minor axis. If the area of the triangle $FBF'$ is $\sqrt{3}$ sq. units,then the eccentricity of the ellipse is

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

If tangents are drawn from any point on the circle $x^2+y^2=25$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$,then the angle between the tangents 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