Question

An ellipse has eccentricity $\frac{1}{2}$ and one focus at $S \left(\frac{1}{2}, 1\right)$. Its one directrix is common tangent (nearer to $S$ ) to the circle $x^2+y^2=1$ and $x^2-y^2=1$. The equation of the ellipse is

• A. $9\left(x-\frac{1}{3}\right)^2+12(y-1)^2=1$
• B. $12\left(x-\frac{1}{3}\right)^2+9(y-1)^2=1$
• C. $3\left(x-\frac{1}{3}\right)^2+4(y-1)^2=1$
• D. $9\left(x-\frac{1}{3}\right)^2+16(y-1)^2=1$

Answer 1

Answer: (A)

Directrix $\Rightarrow x =1$
$\text { focus }=\left(\frac{1}{2}, 1\right) $
$e =\frac{1}{2} $
$\Theta \frac{ PS }{ PM }= e \Rightarrow PS ^2= e ^2 PM ^2$
$\Rightarrow \left( x -\frac{1}{2}\right)^2+( y -1)^2=\left(\frac{1}{2}\right)^2( x -1)^2 $
$\Rightarrow 9\left( x -\frac{1}{3}\right)^2+12( y -1)^2=1 $