The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ is -

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

Find the position of the point $(4, -3)$ with respect to the ellipse $2x^2 + 5y^2 = 20$.

Consider an ellipse with foci at $(5, 15)$ and $(21, 15)$. If the $X$-axis is a tangent to the ellipse,then the length of its major axis equals

If an ellipse with foci at $(3,3)$ and $(-4,4)$ is passing through the origin,then the eccentricity of that ellipse is

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 $......$