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 at the point $P$, to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1$. The equation of the ellipse is 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. None of these

Answer 1

Answer: (B)

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

Clearly, the common tangent to the circle $x^2+y^2=1$ and hyperbola $x^2-y^2=1$ is x=1$

$[$ which is nearer to $P(\frac{1}{2},1)].$

Given, one focus at $P(\frac{1}{2},1).$

$\therefore$ equation of the directrix is $x=1.$

$\therefore$ ellipse is $\sqrt{{(x-\frac{1}{2})}^2+{(y-1)}^2}=\frac{1}{2}(x-1)$

On simplification, it becomes $9{(x-\frac{1}{3})}^2+12{(y-1)}^2=1.$