Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$

Since the area of ellipse is $A=\pi ab\Rightarrow A^2=\pi^2a^2b^2$ and the circle and the ellipse touch each other internally. This much I can visualise, but how to find out $a$ and $b$ from this?

Means ellipse and circle touches each other. So $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and $(x-1)^2+y^2 = 1$
So Ellipse and Circle touches each other. So we will solve these two equations.
$\displaystyle \frac{x^2}{a^2}+\frac{1-(x-1)^2}{b^2} = 1\Rightarrow b^2x^2+a^2\left[1-(x-1)^2\right] = a^2b^2$
so $\displaystyle b^2x^2+a^2\left[1-(x^2+1-2x)\right]=a^2b^2$
So $\displaystyle b^2x^2+a^2-a^2x^2-a^2+2a^2x-a^2b^2 =0$
So $(b^2-a^2)x^2+2a^2x-a^2b^2$
So Discriminat of above equation is $=0$(Bcz Courve touches each other)
So $\displaystyle 4a^4+4(b^2-a^2)\cdot (a^2b^2) =0$
So $\displaystyle a^2+(b^2-a^2)b^2=0\Rightarrow a^2+b^4-a^2b^2=0$
