The equation of the tangent to the ellipse $x^2+16y^2=16$ which makes an angle $60^{\circ}$ with the $X$-axis is

The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ is -

If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4y^2=36$ is $r$,then $12r^2$ is equal to

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

For the ellipse $3x^2 + 4y^2 = 12$,the length of the latus rectum is:

On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$,let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x+2y=0$. Let $S$ and $S'$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SPS'$,then the value of $(5-e^{2}) \cdot A$ is: