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

Answer 1

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

Given
Coordinate of focus $\equiv (-1,1)$
Equation of directrix: $x-y+3=0$
Eccentricity $=\frac{1}{2}$
We know that, an ellipse is the locus of the point which moves in the plane so that ratio of its distance from a focus and a directrix is equal to eccentricity of the ellipse.
$\therefore$ From definition of the ellipse
$\frac{PS}{PM}=e$
$\Rightarrow \frac{\sqrt{{(x+1)}^2+{(y-1)}^2}}{\frac{x-y+3}{\sqrt{2}}}=\frac{1}{2}$
$\Rightarrow \sqrt{{(x+1)}^2+{(y-1)}^2}=\frac{1}{2}\frac{x-y+3}{\sqrt{2}}$
By squaring both side we get
${(x+1)}^2+{(y-1)}^2=\frac{1}{8}{(x-y+3)}^2$