Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

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 $P(\alpha, \beta)$ is a point on the curve $9x^2 + 4y^2 = 144$ in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at $P$ with the coordinate axes is $S$,then

One of the foci of an ellipse is $(2,-3)$ and its corresponding directrix is $2x+y=5$. If the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$,then the coordinates of the other focus are

The equations of the latus rectum of the ellipse $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ are

The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on: