Question

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

• A. $7x^2+2xy+7y^2-10x+10y+7=0$
• B. $7x^2-2xy+7y^2-10x-10y+7=0$
• C. $7x^2+2xy+7y^2+10x-10y+7=0$
• D. $7x^2-2xy+7y^2+10x+10y-7=0$

Answer 1

The correct option is C $7x^2+2xy+7y^2+10x-10y+7=0$

If $P(x,y)$ be any point on the ellipse, S be its focus and PN be the perpendicular from P on directrix, then by definition of the ellipse ${PS}^2=e^2{PN}^2$

Hence ${(x+1)}^2+{(y-1)}^2=\frac{1}{4}{\left(\frac{x-y+3}{\sqrt{2}}\right)}^2=\frac{(x-y+3{)}^2}{8}$

As focus is $(-1,1)$ and directrix is $x-y+3=0$

$\Rightarrow 8(x^2+y^2+2x-2y+2)=x^2+y^2+9-2xy-6y+6x$

$\Rightarrow {7x}^2+2xy+{7y}^2+10x-10y+7=0$