The equation of an ellipse in its standard form,given the distance between its foci is $2$ units and the length of its latus rectum is $\frac{15}{2}$ units,is

If the coordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic $9x^{2} + 16y^{2} = 144$,then $PA + PB$ is equal to

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b < a)$ be an ellipse with major axis $AB$ and minor axis $CD$. Let $F_1$ and $F_2$ be its two foci,with $A, F_1, F_2, B$ in that order on the segment $AB$. Suppose $\angle F_1CB = 90^{\circ}$. The eccentricity of the ellipse is:

The position of the point $(4, -3)$ with respect to the ellipse $2x^2 + 5y^2 = 20$ is

$A$ vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $PQR$,$\Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$,then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

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