If $F_1$ and $F_2$ are the foci of the ellipse $16 x^2+25 y^2=400$ and $P$ is any point on it,then the value of the product $P F_1 \cdot P F_2$ lies in the interval

For an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ with vertices $A$ and $A'$,a tangent drawn at the point $P$ in the first quadrant meets the $y$-axis at $Q$,and the chord $A'P$ meets the $y$-axis at $M$. If $O$ is the origin,then $OQ^2 - MQ^2$ is equal to:

Consider an ellipse,whose centre is at the origin and its major axis is along the $x-$ axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$,then the area (in sq. units) of the quadrilateral inscribed in the ellipse,with the vertices as the vertices of the ellipse,is

Let $E$ be an ellipse whose axes are parallel to the coordinate axes,having its center at $(3, -4)$,one focus at $(4, -4)$,and one vertex at $(5, -4)$. If $mx - y = 4$ with $m > 0$ is a tangent to the ellipse $E$,then the value of $5m^{2}$ is equal to .....

The values that $m$ can take,so that the straight line $y=4x+m$ touches the curve $x^2+4y^2=4$ are

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is: