Find the equation of the ellipse whose foci are $(\pm 2, 0)$ and eccentricity is $1/2$.

$S=(-1, 1)$ is the focus,$2x-3y+1=0$ is the directrix corresponding to $S$,and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse,then $3a+2b=$

The line $x = 8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4\sqrt{3})$ and intersects the $x$-axis at $Q$,then $(3PQ)^2$ is equal to $........$

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval: