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 

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $ T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ Then $n ( S \cap T )$ is equal to $......$

Find the coordinates of the foci, the rertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$

The equation of the ellipse whose one focus is at $(4, 0)$ and whose eccentricity is $4/5$, is

What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

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