Question

If Y-axis is the directrix of the ellipse with eccentricity $e=1/2$ and corresponding focus is at $(3,0)$, then the equation to its auxiliary circle is

• A. $x^2+y^2-8x+12=0$
• B. $x^2+y^2-8x-12=0$
• C. $x^2+y^2-8x+9=0$
• D. $x^2+y^2=4$

Answer 1

The correct option is A $x^2+y^2-8x+12=0$

Let center of the ellipse be $(h,k)$

Given that y-axis is the directrix.

$\Rightarrow x=0$

We know that directrix is $x=h-\frac{a}{e}$

So, $\Rightarrow h-\frac{a}{e}=0$

$\Rightarrow h=\frac{a}{e}$

$\Rightarrow h=\frac{a}{1/2}$

$\Rightarrow h=2a$

We know that focus $=(h-ae,k)$

$\Rightarrow (h-ae,k)=(3,0)$

$\Rightarrow h-ae=3$

$\Rightarrow 2a-\left(\frac{a}{2}\right)=3$

$\Rightarrow \frac{3a}{2}=3$

$\Rightarrow a=2$

$\Rightarrow a^2=4$

We know that $b^2=a^2(1-e^2)$

$\Rightarrow h=2a=4;k=0$

$\Rightarrow b^2=4(1-\frac{1}{4})$

$\Rightarrow b^2=3$

Thus equation of ellipse$\Rightarrow \frac{(x-4{)}^2}{4}+\frac{y^2}{3}=1$

Hence, equation of auxiliary circle is $\Rightarrow (x-h{)}^2+(y-k{)}^2=a^2$

$\Rightarrow (x-4{)}^2+y^2=4$

$\Rightarrow x^2+y^2-8x+16=4$

$\Rightarrow x^2+y^2-8x+12=0$