Question

An ellipse has eccentricity $\frac{1}{2}$ and focus at the point $P \left(\frac{1}{2}, 1\right)$ 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 in standard form is

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

Answer 1

Answer: (B)

Common tangent to the circle $x^2+y^2=1$
and Hyperbola $x^2-y^2=1$
is $x =1$
Point $P \left(\frac{1}{2}, 1\right)$, equation of the directrix $\Rightarrow x =1$
Ellipse : $PS = e \cdot pm$
$\sqrt{\left(x-\frac{1}{2}\right)^2+(y-1)^2}=\frac{1}{2}(x-1) $
$\left(x-\frac{1}{2}\right)^2+(y-1)^2=\frac{1}{4}(x-1)^2$
After simplification
$9\left( x -\frac{1}{3}\right)^2+12( y -1)^2=1 $