The length of the latus rectum of the ellipse $5{x^2} + 9{y^2} = 45$ 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

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

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3 r )^{2}$ is equal to

If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is

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