Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P, Q$ and $Q^{\prime}$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M \left( P , Q ^{\prime}\right)$ be the mid-point of the line segment joining $P$ and $Q ^{\prime}$. Then the maximum possible value of the distance between $M ( P , Q )$ and $M \left( P , Q ^{\prime}\right)$, as $P, Q$ and $Q^{\prime}$ vary on $E$, is. . . . .
$2$
$3$
$4$
$5$
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$ . If $L$ is also a tangent to the ellipse $\frac{ x ^{2}}{2}+\frac{ y ^{2}}{ b }=1,$ then the value of $b$ is equal to ..... .
The locus of point of intersection of two perpendicular tangent of the ellipse $\frac{{{x^2}}}{{{9}}} + \frac{{{y^2}}}{{{4}}} = 1$ is :-
Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to :
Define the collections $\left\{ E _1, E _2, E _3, \ldots ..\right\}$ of ellipses and $\left\{ R _1, K _2, K _3, \ldots ..\right\}$ of rectangles as follows : $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$
$K _1$ : rectangle of largest area, with sides parallel to the axes, inscribed in $E _1$;
$E_n$ : ellipse $\frac{x^2}{a_n^2}+\frac{y^2}{b_{n}^2}=1$ of largest area inscribed in $R_{n-1}, n>1$;
$R _{ n }$ : rectangle of largest area, with sides parallel to the axes, inscribed in $E _{ n }, n >1$.
Then which of the following options is/are correct?
$(1)$ The eccentricities of $E _{18}$ and $E _{19}$ are NOT equal
$(2)$ The distance of a focus from the centre in $E_9$ is $\frac{\sqrt{5}}{32}$
$(3)$ The length of latus rectum of $E_Q$ is $\frac{1}{6}$
$(4)$ $\sum_{n=1}^N\left(\right.$ area of $\left.R_2\right)<24$, for each positive integer $N$
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 $36 x^{2}+4 y^{2}=144$