Let $S = 0$ is an ellipse whose vartices are the extremities of minor axis of the ellipse $E:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,a > b$ If $S = 0$ passes through the foci of $E$ , then its eccentricity is (considering the eccentricity of $E$ as $e$ )
$\sqrt {\frac{{1 - 2{e^2}}}{{1 - {e^2}}}} $
$\frac{1}{{\sqrt {1 + {e^2}} }}$
$\frac{{1 - 2{e^2}}}{{1 - {e^2}}}$
$\frac{{{e^2}}}{{1 + {e^2}}}$
Let $L$ is distance between two parallel normals of , $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,\,a > b$ then maximum value of $L$ is
Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP ^{2}+ SQ ^{2}$ is equal to.
The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x-$ axis at $Q$ and $R,$ respectively. Then the area(in sq. units) of the triangle $PQR$ is
The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt 2$ is-