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

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$,be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$,then $a^{2}+b^{2}$ is equal to

The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt{2}$ is-

Let a line $L$ pass through the point of intersection of the lines $bx + 10y - 8 = 0$ and $2x - 3y = 0$,where $b \in R - \{\frac{4}{3}\}$. If the line $L$ also passes through the point $(1, 1)$ and touches the circle $17(x^2 + y^2) = 16$,then the eccentricity of the ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ is:

Let $S_{1}: x^{2}+y^{2}=9$ and $S_{2}:(x-2)^{2}+y^{2}=1$. Then the locus of the center of a variable circle $S$ which touches $S_{1}$ internally and $S_{2}$ externally always passes through the points :

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to $......$