Maximum length of chord of the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is 

An ellipse is drawn by taking a diameter of the circle ${\left( {x - 1} \right)^2} + {y^2} = 1$ as its semi-minor axis and a diameter of the circle ${x^2} + {\left( {y - 2} \right)^2} = 4$ is semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :

The equations of the tangents of the ellipse $9{x^2} + 16{y^2} = 144$ which passes through the point $(2, 3)$ is

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 $A F_1$ and $A F_2$, when extended, cut the ellipse again at points $B$ and $C$ respectively. Suppose that the $\triangle A B C$ is equilateral. Then, the eccentricity of the ellipse is

Let a line $L$ pass through the point of intersection of the lines $b x+10 y-8=0$ and $2 x-3 y=0$, $b \in R -\left\{\frac{4}{3}\right\}$. If the line $L$ also passes through the point $(1,1)$ and touches the circle $17\left( x ^{2}+ y ^{2}\right)=16$, then the eccentricity of the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{b^{2}}=1$ is.