Question

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

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

Answer 1

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

These are two common tangents to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1$.

There are $x=1$ and $x=-1$

Out of these, $x=1$ is nearer to the point $P(\frac{1}{2},1)$.

Thus, a directrix of the required ellipse is $x=1$.

If $Q(x,y)$ is any point on the ellipse, then its distance from the focus is $QP=\sqrt{{(x-\frac{1}{2})}^2+{(y-1)}^2}$ and its distance from the directrix $x=1$ is $|x-1|$

By definition os ellipse, $QP=e|x-1|$

$\Rightarrow \sqrt{{(x-\frac{1}{2})}^2+{(y-1)}^2}=\frac{1}{2}|x-1|\Rightarrow 3x^2-2x+4y^2-8y+4=0$

$\Rightarrow \frac{{(x-\frac{1}{3})}^2}{\frac{1}{9}}+\frac{{(y-1)}^2}{\frac{1}{12}}=1$