The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with the coordinate axes is

For real numbers $a, b$ $(a > b > 0)$,let $\text{Area} \{(x, y) : x^{2} + y^{2} \leq a^{2} \text{ and } \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \geq 1\} = 30\pi$ and $\text{Area} \{(x, y) : x^{2} + y^{2} \geq b^{2} \text{ and } \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1\} = 18\pi$. Then the value of $(a - b)^{2}$ is equal to

If $B$ is the end point of the minor axis of the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$ $(a > b)$ and $S$ and $S^{\prime}$ are the foci of the ellipse such that $\Delta SBS^{\prime}$ is an equilateral triangle,then the eccentricity $e$ is:

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

An ellipse,with foci at $(0, 2)$ and $(0, -2)$ and minor axis of length $4$,passes through which of the following points?

Let $A=\{(\alpha, \beta) \in R \times R :|\alpha-1| \leq 4 \text{ and }|\beta-5| \leq 6\}$ and $B=\left\{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}$. Then