If $x+\sqrt{3} y=3$ is the tangent to the ellipse $2 x^2+3 y^2=k$ at a point $P$,then the equation of the normal to this ellipse at $P$ is

  • A
    $5 x-2 \sqrt{3} y=1$
  • B
    $x-\sqrt{3} y=2$
  • C
    $x-\sqrt{3} y+1=0$
  • D
    $3 x-\sqrt{3} y=1$

Explore More

Similar Questions

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
$(A) e_1^2+e_2^2=\frac{43}{40}$
$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C) |e_1^2-e_2^2|=\frac{5}{8}$
$(D) e_1 e_2=\frac{\sqrt{3}}{4}$

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to

$A$ particle is travelling in a clockwise direction on the ellipse $\frac{x^2}{100} + \frac{y^2}{25} = 1$. If the particle leaves the ellipse at the point $(-8, 3)$ and travels along the tangent to the ellipse at that point,then the point where the particle crosses the $Y$-axis is:

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse with foci $F_1$ and $F_2$. Let $AO$ be its semi-minor axis,where $O$ is the centre of the ellipse. The lines $AF_1$ and $AF_2$,when extended,cut the ellipse again at points $B$ and $C$ respectively. Suppose that the $\triangle ABC$ is equilateral. Then,the eccentricity of the ellipse 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