Locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is
${({x^2} + {y^2})^2} = {b^2}{x^2} + {a^2}{y^2}$
${({x^2} + {y^2})^2} = {b^2}{x^2} - {a^2}{y^2}$
${({x^2} + {y^2})^2} = {a^2}{x^2} - {b^2}{y^2}$
${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$
Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $ $ \&$ the diameter of the inscribed circle of triangle $OCF$ is $2, $ then the product $ (AB)\,(CD) $ is equal to
The length of the latus rectum of an ellipse is $\frac{1}{3}$ of the major axis. Its eccentricity is
The equation of an ellipse whose focus $(-1, 1)$, whose directrix is $x - y + 3 = 0$ and whose eccentricity is $\frac{1}{2}$, is given by
For some $\theta \in\left(0, \frac{\pi}{2}\right),$ if the eccentricity of the hyperbola, $x^{2}-y^{2} \sec ^{2} \theta=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec ^{2} \theta+y^{2}=5,$ then the length of the latus rectum of the ellipse is
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.