An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.

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

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

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

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

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

The  the circle passing through the foci of the $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$ and having centre at $(0,3) $ is

If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to: