Consider the parabola $P : y^2 = 4x$ and the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of $P$ and $E$ be their common latus rectum. If the eccentricity of $E$ is $e$, then $e^2 + 2\sqrt{2}$ is equal to . . . . . .

The eccentricity of an ellipse,with its centre at the origin,is $\frac{1}{2}$. If one of the directrices is $x = 4$,then the equation of the ellipse is

The equation of the ellipse whose latus rectum is $8$ and whose eccentricity is $\frac{1}{\sqrt{2}}$,referred to the principal axes of coordinates,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 quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is .............. $sq. \text{ units}$.

The eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are given by