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 

If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4 y^2=36$ is $r$, then $12 r^2$ is equal to

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 :

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 $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$

Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a ponit $\mathrm{P}$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta),$ then $\beta$ is equal to

A chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends right angle at its  centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-