The locus of the poles of normal chords of an ellipse is given by

  • A

    $\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {({a^2} - {b^2})^2}$

  • B

    $\frac{{{a^3}}}{{{x^2}}} + \frac{{{b^3}}}{{{y^2}}} = {({a^2} - {b^2})^2}$

  • C

    $\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {({a^2} + {b^2})^2}$

  • D

    $\frac{{{a^3}}}{{{x^2}}} + \frac{{{b^3}}}{{{y^2}}} = {({a^2} + {b^2})^2}$

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 $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)

$(A)$ $e_1^2+e_2^2=\frac{43}{40}$

$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$

$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$

$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$

  • [IIT 2015]

On the ellipse $\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{8} = 1$ the point $M$ nearest to the line $2x - 3y + 25 = 0$ is

If $3 x+4 y=12 \sqrt{2}$ is a tangent to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$ for some a $\in \mathrm{R},$ then the distance between the foci of the ellipse is

  • [JEE MAIN 2020]

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}}} = $

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$, is .............. $\mathrm{sq. \,units}$

  • [IIT 2003]