Question

The equation of the ellipse, whose focus is the point $(-1,1)$, whose directrix is the straight line $x-y+3=0$ and whose eccentricity is $\frac{1}{2}$ is :

• A. $(x+1{)}^2+(y-1{)}^2=\frac{1}{2}(x-y+3{)}^2$
• B. $(x+1{)}^2+(y-1{)}^2=\frac{1}{8}(x-y+3{)}^2$
• C. $(x+1{)}^2+(y-1{)}^2=\frac{1}{8}(x-y+1{)}^2$
• D. $(x+1{)}^2+(y-1{)}^2=\frac{1}{6}(x-y+3{)}^2$

Answer 1

The correct option is B $(x+1{)}^2+(y-1{)}^2=\frac{1}{8}(x-y+3{)}^2$

Let $P(x,y)$ be a point on the ellipse.
So, $\frac{\text{distance between P and focus}}{\text{distance from P to the directrix}}=e=\frac{1}{2}$

$\Rightarrow \frac{\sqrt{(x+1{)}^2+(y-1{)}^2}}{\left|\frac{x-y+3}{\sqrt{2}}\right|}=e=\frac{1}{2}$

On squaring both the sides, we get
$(x+1{)}^2+(y-1{)}^2=\frac{1}{8}(x-y+3{)}^2$