In an ellipse,two vertices are $(5,0)$ and $(0,-4)$. Then the equation of the ellipse is

The ellipse $x^2 + 4y^2 = 4$ is inscribed in a rectangle whose sides are parallel to the coordinate axes. Find the equation of the ellipse that circumscribes this rectangle.

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

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:

If the equations $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$ for $0 \leq \theta < 2 \pi$ represent an ellipse,then the point of intersection of the normal drawn at $P(\theta = \pi/4)$ to this ellipse and its major axis is:

If the maximum distance of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$,where $b < 2$,from the origin is $1$,then the eccentricity of the ellipse is: