If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6, 0)$,foci $(\pm 4, 0)$.

Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$, then $\alpha^2+\beta^2$ is equal to:

The maximum length of a chord of the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,such that the eccentric angles of its extremities differ by $\frac{\pi}{2}$,is:

Let $A(\alpha, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$. Let the point $P$ divide the line segment $AB$ internally in the ratio $7:3$. Let $3x - 25 = 0$ be a directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$. If from $S$,the perpendicular on the $x$-axis passes through $P$,then the length of the latus rectum of $E$ is equal to

The eccentricity of the ellipse $4x^2 + 9y^2 = 36$ is