The locus of point of intersection of two perpendicular tangent of the ellipse  $\frac{{{x^2}}}{{{9}}} + \frac{{{y^2}}}{{{4}}} = 1$ is :-

The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $P Q$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3 r )^{2}$ is equal to

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is

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)

$(A)$ $e_1^2+e_2^2=\frac{43}{40}$

$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$

$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$

$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$

If a tangent having slope of $ - \frac{4}{3}$ to the ellipse $\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{{32}} = 1$ intersects the major and minor axes in points $A$ and $B$ respectively, then the area of $\Delta OAB$ is equal to .................. $\mathrm{sq. \, units}$ ($O$ is centre of the ellipse)