The equation $\frac{x^2}{10-a} + \frac{y^2}{4-a} = 1$ represents an ellipse when:

The area (in square units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is:

If the foci of an ellipse are $(\pm \sqrt{5}, 0)$ and its eccentricity is $\frac{\sqrt{5}}{3}$,then the equation of the ellipse is

Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e = \frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle,whose vertex $P$ is on the circle $C$ and the side $QR$ of length $2$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ is:

The area of the triangle formed by the tangent and normal to the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the point $P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ with the $x$-axis is

If $S$ and $S'$ are the two foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a < b$,and $P(x_1, y_1)$ is a point on the ellipse,then $SP + S'P = \dots$