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

Let a tangent drawn at any point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut the $X$-axis at $Q$. Let $R$ be the image of $Q$ with respect to $y=x$. If $S$ is a circle with $QR$ as its diameter,then the fixed point through which the circle $S$ passes is

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

The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse $9x^2 + 4y^2 = 72$ at the point $(2, 3)$ with the $X$-axis 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

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is