If the line $\frac{x}{a} + \frac{y}{b} = \sqrt{2}$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then its eccentric angle $\theta = ............^{\circ}$

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 $\alpha^2+\beta^2=$

The line $y = 2t^2$ intersects the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ in real points if

The length of the latus rectum of the ellipse $5x^2 + 9y^2 = 45$ is

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

The eccentric angle of the point $P(-6, 2)$ on the ellipse $\frac{x^2}{48} + \frac{y^2}{16} = 1$ is: (in $^{\circ}$)