The eccentricity of the ellipse $\frac{{{{(x - 1)}^2}}}{9} + \frac{{{{(y + 1)}^2}}}{{25}} = 1$ is

Let the foci and length of the latus rectum of an ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ be $( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1$ equals

If $P$ lies in the first quadrant on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ (where $a > b$ ), and tangent & normal drawn at $P$ meets major axis at the points $T$ & $N$ respectively, then the value of $\frac{{\left( {\left| {{F_2}N} \right| + \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| - \left| {{F_1}T} \right|} \right)}}{{\left( {\left| {{F_2}N} \right| - \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| + \left| {{F_1}T} \right|} \right)}}$ is equal to (where $F_1$ & $F_2$ are the foci $(ae, 0)$ & $(-ae, 0)$ respectively)

The equation of tangent and normal at point $(3, -2)$ of ellipse $4{x^2} + 9{y^2} = 36$ are

The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9 y^2=9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A, M$ and the origin $O$ is

The line, $ lx + my + n = 0$  will cut the ellipse $\frac{{{x^2}}}{{{a^2}}}$ $+$ $\frac{{{y^2}}}{{{b^2}}}$ $= 1 $ in points whose eccentric angles differ by $\pi /2$  if :