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

Tangents at extremities of latus rectum of ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq.\ units$) -

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is

Let $P$ be a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with foci ${F_1}$ and ${F_2}$. If $A$ is the area of the triangle $P{F_1}{F_2}$, then maximum value of $A$ is

If any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ cuts off intercepts of length $h$ and $k$ on the axes, then $\frac{{{a^2}}}{{{h^2}}} + \frac{{{b^2}}}{{{k^2}}} = $

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $  $ \&$  the diameter of the inscribed circle of triangle $OCF$  is $2, $ then the product $ (AB)\,(CD) $ is equal to