The line, $ lx + my + n = 0$ will cut the ellipse $\frac{{{x^2}}}{{{a^2}}}$ $+$ $\frac{{{y^2}}}{{{b^2}}}$ $= 1 $ in points whose eccentric angles differ by $\pi /2$ if :
$a^2l^2 + b^2n^2 = 2 m^2$
$a^2m^2 + b^2l^2 = 2 n^2$
$a^2l^2 + b^2m^2 = 2 n^2$
$a^2n^2 + b^2m^2 = 2 l^2$
Let the line $2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0, \mathrm{k}>0$, intersect the $\mathrm{x}$-axis and $\mathrm{y}$-axis at the points $\mathrm{A}$ and $\mathrm{B}$, respectively. If the equation of the circle having the line segment $\mathrm{AB}$ as a diameter is $\mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}-2 \mathrm{y}=0$ and the length of the latus rectum of the ellipse $\mathrm{x}^2+9 \mathrm{y}^2=\mathrm{k}^2$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime, then $2 \mathrm{~m}+\mathrm{n}$ is equal to
The area (in sq, units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$ is :
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
The latus rectum of an ellipse is $10$ and the minor axis is equal to the distance between the foci. The equation of the ellipse is