If $3 x+4 y=12 \sqrt{2}$ is a tangent to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$ for some a $\in \mathrm{R},$ then the distance between the foci of the ellipse is

An ellipse is inscribed in a circle and a point is inside a circle is choosen at random. If the probability that this point lies outside the ellipse is $\frac {2}{3}$ then eccentricity of ellipse is $\frac{{a\sqrt b }}{c}$ . Where $gcd( a, c) = 1$ and $b$ is square free integer ($b$ is not divisible by square of any integer except $1$ ) then $a · b · c$ is

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x = 4$, then the equation of the ellipse is

The angle of intersection of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and circle ${x^2} + {y^2} = ab$, is

If the normal at any point $P$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ meets the co-ordinate axes in $G$ and $g$ respectively, then $PG:Pg = $