What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$  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 :

The angle of intersection of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and circle ${x^2} + {y^2} = ab$, is

The locus of mid points of parts in between axes and tangents of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ will be