The coordinates of the foci of the ellipse $16x^{2} + 9y^{2} = 144$ are

Find the equation for the ellipse that satisfies the given conditions: Length of major axis $= 26$,foci $= (\pm 5, 0)$.

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
$(A) e_1^2+e_2^2=\frac{43}{40}$
$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C) |e_1^2-e_2^2|=\frac{5}{8}$
$(D) e_1 e_2=\frac{\sqrt{3}}{4}$

If $B$ is the end point of the minor axis of the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$ $(a > b)$ and $S$ and $S^{\prime}$ are the foci of the ellipse such that $\Delta SBS^{\prime}$ is an equilateral triangle,then the eccentricity $e$ is:

The equation $\frac{x^2}{2 - r} + \frac{y^2}{r - 5} + 1 = 0$ represents an ellipse,if

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse with foci $F_1$ and $F_2$. Let $AO$ be its semi-minor axis,where $O$ is the centre of the ellipse. The lines $AF_1$ and $AF_2$,when extended,cut the ellipse again at points $B$ and $C$ respectively. Suppose that the $\triangle ABC$ is equilateral. Then,the eccentricity of the ellipse is