An ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is tangent to each of the circles ${(x - 2)}^{2} + y^{2} = 4$ and ${(x + 2)}^{2} + y^{2} = 4$ at two points. If the area of the ellipse is minimized, then which of the following is/are CORRECT ?

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Minimum area of the ellipse is $6 \sqrt{3} π$

The value of $\frac{a^{2}}{b^{2}}$ is 3

The value of $\frac{a^{2}}{b^{2}}$ is $\frac{1}{3}$

Minimum area of the ellipse is $3 \sqrt{3} π$

answer is C, D.

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Detailed Solution

Solving the ellipse and the first circle $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, {(x - 2)}^{2} + y^{2} = 4$
Substitute $y^{2} = 4 - {(x - 2)}^{2}$ in the ellipse equation, we get $(b^{2} - a^{2}) x^{2} + 4 a^{2} x - a^{2} b^{2} = 0$
This quadratic equation has two equal roots
$16 a^{4} - 4 (b^{2} - a^{2}) (- a^{2} b^{2}) = 0 4 a^{2} + b^{4} - a^{2} b^{2} = 0 a^{2} = \frac{b^{4}}{b^{2} - 4}$

Area of ellipse = $π ab = π * \frac{b^{2}}{\sqrt{b^{2} - 4}} * b = \frac{π b^{3}}{\sqrt{b^{2} - 4}} = A (b)$
$\frac{d A}{d b} = 0$ for minimum area
Solving that we get $b^{2} = 6$ and $a^{2} = 18$