Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ is

Let $E$ be the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ and $C$ be the circle ${x^2} + {y^2} = 9$. Let $P$ and $Q$ be the points $(1, 2)$ and $(2, 1)$ respectively. Then

If the line $x\cos \alpha + y\sin \alpha = p$ be normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then

In an ellipse $9{x^2} + 5{y^2} = 45$, the distance between the foci is

Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)$. If $\mathrm{e}$ and $1$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$, then $8 e^2+1^2$ is equal to.

Let $P$ is any point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ . $S_1$ and $S_2$ its foci then maximum area of $\Delta PS_1S_2$ is (in square units)