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