Find the equation of the ellipse with its center at the origin,passing through the points $(-3, 1)$ and $(2, -2)$,given that $a > b$.

If the straight line $8x + 3\sqrt{2}y = 36$ touches the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 2$ at $(a, b)$,then $a + \sqrt{2}b =$

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

Let $A = \{(x, y) : y = mx + 1\}$,$B = \{(x, y) : x^2 + 4y^2 = 1\}$,and $C = \{(\alpha, \beta) : (\alpha, \beta) \in A \text{ and } (\alpha, \beta) \in B \text{ and } \alpha > 0\}$. If set $C$ is a singleton set,then the sum of all possible values of $m$ is:

The area of the quadrilateral formed by drawing tangents at the ends of the latus recta of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is

For $0 < \theta < \frac{\pi}{2}$,four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents,the minimum value of $A(\theta)$ is