The foci of $16{x^2} + 25{y^2} = 400$ are

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

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 $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

If tangents are drawn from point $P(3\ sin\theta + 4\ cos\theta , 3\ cos\theta\ -\ 4\ sin\theta)$ , $\theta = \frac {\pi}{8}$ to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$ then angle between the tangents is

The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes.

Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ is