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

The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is

Extremities of the latera recta of the ellipses $\frac{{{x^2}}}{{{a^2}}}\,\, + \,\,\frac{{{y^2}}}{{{b^2}}}\, = \,1\,$ $(a > b)$  having a given major axis $2a$  lies on

If the normal at any point $P$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ meets the co-ordinate axes in $G$ and $g$ respectively, then $PG:Pg = $

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 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