The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is such that it has the least area but contains the circle $(x-1{)}^2+y^2=1$ then the length of latus rectum of ellipse is equal to

B
Solving these equation $\frac{x^2}{a^2}+\frac{1-(x-1{)}^2}{b^2}=1$
$\begin{array}{r}{b}^2{x}^2+a^2\left[1-(x-1{)}^2\right]=a^2{b}^2\\ \left(b^2-a^2\right)x^2+2a^{2}x-a^2{b}^2=0\end{array}$
For least area circle must touch the ellipse D=0 
$\begin{array}{l}4a^2+4a^2{b}^2\left(b^2-a^2\right)=0\\ a^2+b^2\left(b^2-a^2\right)=0\\ a^2+b^2\left(-a^2{e}^2\right)=0\\ 1-b^2{e}^2=0\\ \Rightarrow b=\frac{1}{e}\\ a^2=\frac{b^2}{1-e^2}=\frac{1}{e^2\left(1-e^2\right)}\end{array}$
$a=\frac{1}{e\sqrt{1-e^2}}$
Area is minimum $f\left(e\right)=e^4-e^6$ is maximum as Area $=\frac{x}{\sqrt{e^4-e^6}},f^{\mathrm{\prime }}\left(e\right)=4e^3-6e^5=0$
$e=\sqrt{\frac{2}{3}}\text{ maximum for }f\left(e\right)\text{ least }e=\sqrt{\frac{2}{3}}$
The equation of ellipse  $=2x^2+6y^2=9$
Length of latus rectum $\frac{2b^2}{a}=\frac{2\cdot \frac{3}{2}}{\frac{3}{\sqrt{2}}}=\sqrt{2}$