Question

An ellipse has eccentricity $\frac{1}{2}$ and one of the foci at the point $S(\frac{1}{2},1).$ One of the directrices of the ellipse is the common tangent to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1,$ corresponding to the focus $S.$ The equation of the ellipse in standard form is

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

Answer 1

The correct option is B $9{(x-\frac{1}{3})}^2+12{(y-1)}^2=1$

Common tangent to the circle $x^2+y^2=1$ and hyperbola $x^2-y^2=1$ is $x=1$ or $x=-1$


Corresponding to $S(\frac{1}{2},1),$ equation of the directrix is $x=1$

By definition of ellipse,
$PS=e\cdot PM$
$\Rightarrow \sqrt{{(x-\frac{1}{2})}^2+{(y-1)}^2}=\frac{1}{2}(x-1)$
$\Rightarrow {(x-\frac{1}{2})}^2+{(y-1)}^2=\frac{1}{4}{(x-1)}^2$
$\Rightarrow 9{(x-\frac{1}{3})}^2+12{(y-1)}^2=1$