Find the equations of the tangents to the ellipse $3x^{2} + 4y^{2} = 12$ which are perpendicular to the line $y + 2x = 4$.

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to

If any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ cuts off intercepts of length $h$ and $k$ on the axes,then $\frac{a^2}{h^2} + \frac{b^2}{k^2} = $

Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $10$. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$,$t \in R$,then $a^2 + b^2$ is equal to:

If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ $(b < 3)$ to its corresponding directrix is $\frac{4}{\sqrt{5}}$,then the slope of the tangent to this ellipse drawn at $\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is

The number of points which can be expressed in the form $(p_1/q_1, p_2/q_2)$,where $p_i$ and $q_i$ $(i = 1, 2)$ are co-primes,and lie on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is: