The distance of the point $'\theta '$on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ from a focus is

Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1<0, y_2<0$, be the end points of the latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are

$(A)$ $x^2+2 \sqrt{3} y=3+\sqrt{3}$

$(B)$ $x^2-2 \sqrt{3} y=3+\sqrt{3}$

$(C)$ $x^2+2 \sqrt{3} y=3-\sqrt{3}$

$(D)$ $x^2-2 \sqrt{3} y=3-\sqrt{3}$

If $F_1$ and $F_2$ be the feet of the perpendicular from the foci $S_1$ and $S_2$ of an ellipse $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{3} = 1$ on the tangent at any point $P$ on the ellipse, then $(S_1 F_1) (S_2 F_2)$ is equal to

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices than the mid points of the tangents intercepted between the coordinate axes lie on the curve

Let the foci and length of the latus rectum of an ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ be $( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1$ equals

$P$ is a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with $AA'$ as the major axis. Then the maximum value of the area of $\Delta APA'$ is