The y-axis is the directrix of the ellipse with eccentricity e=12 and the corresponding focus is at (3,0), equation to its auxiliary circle is:

Directrix: x=0e=12 Focus =(3,0)∴(x−3)2+y2=12.|x| ∴(x−3)2+y2=14.x2 ⇒4(x−3)2+4y2=x2 ⇒(x−4)24+y23=1∴a=2;b=3; centre (4,0)⇒ Auxiliary circle is (x−4)2+y2=4

The y-axis is the directrix of the ellipse with eccentricity e=12 and the corresponding focus is at (3,0), equation to its auxiliary circle is:

Directrix: x=0e=12 Focus =(3,0)∴(x−3)2+y2=12.|x| ∴(x−3)2+y2=14.x2 ⇒4(x−3)2+4y2=x2 ⇒(x−4)24+y23=1∴a=2;b=3; centre (4,0)⇒ Auxiliary circle is (x−4)2+y2=4

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The y-axis is the directrix of the ellipse with eccentricity $e = \frac{1}{2}$ and the corresponding focus is at $(3, 0)$ , equation to its auxiliary circle is:

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$x^{2} + y^{2} = 4$

$x^{2} + y^{2} - 8 x + 9 = 0$

$x^{2} + y^{2} - 8 x + 12 = 0$

$x^{2} + y^{2} - 8 x - 12 = 0$

answer is A.

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Detailed Solution

Directrix: $x = 0$
$e = \frac{1}{2}$
Focus $= (3, 0)$
$∴ \sqrt{{(x - 3)}^{2} + y^{2}} = \frac{1}{2} . | x | ∴ {(x - 3)}^{2} + y^{2} = \frac{1}{4} . x^{2} \Rightarrow 4 {(x - 3)}^{2} + 4 y^{2} = x^{2} \Rightarrow \frac{{(x - 4)}^{2}}{4} + \frac{y^{2}}{3} = 1$