Maximum length of chord of the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is
$4$
$2\sqrt 2 $
$16$
$8$
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)
$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$
The length of the latus rectum of the ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{49}} = 1$
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
Let $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \mathrm{a}\,>\,\mathrm{b} .$ Let $\mathrm{E}_{2}$ be another ellipse such that it touches the end points of major axis of $E_{1}$ and the foci $E_{2}$ are the end points of minor axis of $E_{1}$. If $E_{1}$ and $E_{2}$ have same eccentricities, then its value is :
The locus of the foot of perpendicular drawn from the centre of the ellipse ${x^2} + 3{y^2} = 6$ on any tangent to it is