Three circles, each of radius 1, are packed inside an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0),(2\theta =\angle ABC),$
 as shown in the fig. Two of these circles, touch the ellipse at two points each, and the middle circle touches it at the end of minor axis. A, B and C are centres of these circles. 

If the area of ellipse is minimum, then which of the following options is/are correct?  
 

$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$area,\Delta =\pi abOBJ:\Delta =f\left(\theta \right) \therefore b-2\cos \theta =1\Rightarrow b=1+2\cos \theta$

Also normal to ellipse at H passes through C 

$LetH(p,q)\Rightarrow N@H\equiv \frac{a^{2}x}{p}-\frac{b^{2}y}{q}=a^2-b^2=a^2{e}^2\Rightarrow c(p{e}^2,0)$

$p{e}^2=2\sin \theta andb-1=2\cos \theta \therefore {\left(p{e}^2\right)}^2+{(b-1)}^2=4$

$HC=1\therefore {(p-2\sin \theta )}^2+q^2=1$

$HC=1\therefore {(p-2\sin \theta )}^2+q^2=1\therefore {\left(p(1-e^2)\right)}^2+q^2=1$

$alsoHliesonellipse\Rightarrow \frac{p^2}{a^2}+\frac{q^2}{b^2}=1\Rightarrow p^2(1-e^2)+q^2=b^2$

$p^2{e}^2(1-e^2)=b^2-1and{p}^2{e}^4=4-{(b-1)}^2$

$\Rightarrow \frac{1-e^2}{e^2}=\frac{b^2-1}{3-b^2+2b}\Rightarrow 1-e^2=\frac{b^2-1}{2+2b}=\frac{b^2}{a^2}$

$\Rightarrow a^2=\frac{2b^2}{b-1}\Rightarrow \Delta =\pi ab=\frac{\pi b^2\sqrt{2}}{\sqrt{b-1}}=\frac{\pi \sqrt{2}}{\sqrt{b^{-3}-b^{-4}}}$

$\therefore \frac{d\Delta }{db}=0\Rightarrow b=\frac{4}{3}\Rightarrow 2\cos \theta =\frac{1}{3}\Rightarrow \tan \theta =\sqrt{35}$