The length of the latus rectum of the ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{49}} = 1$

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

From the point$ C(0,\lambda )$ two tangents are drawn to ellipse $x^2\ +\ 2y^2\ = 4$ cutting major axis at $A$ and $B$. If  area of $\Delta$ $ABC$ is minimum, then value of $\lambda$  is-

Let $P \left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q , R$ and $S$ be four points on the ellipse $9 x^2+4 y^2=36$. Let $P Q$ and $RS$ be mutually perpendicular and pass through the origin. If $\frac{1}{( PQ )^2}+\frac{1}{( RS )^2}=\frac{ p }{ q }$, where $p$ and $q$ are coprime, then $p+q$ is equal to $.........$.

For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ 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 $\frac{x^{2}}{36}+\frac{y^2} {16}=1$