Question

The focus of the parabola $x^2+y^2+2xy-6x-2y+3=0$ is

• A. $(-1,1)$
• B. $(1,1)$
• C. $(1,-\frac{3}{4})$
• D. $(3,1)$

Answer 1

The correct option is B $(1,1)$

The equation of the parabola can be written as
$(x+y{)}^2=6x+2y-3$
Write the above equation as
$(x+y+\lambda {)}^2=2(\lambda +3)x+2(\lambda +1)y+({\lambda }^2-3)$
Choose $\lambda$ such that the lines $x+y+\lambda =0$ and
$2(\lambda +3)x+2(\lambda +1)y+({\lambda }^2-3)=0$ are perpendicular.

$\therefore (-1)\{-\frac{2(\lambda +3)}{2(\lambda +1)}\}=-1$
$\Rightarrow \lambda +3=-\lambda -1\Rightarrow \lambda =-2$

$\therefore$ The equation becomes $(x+y-2{)}^2=2x-2y+1$
Put $X=\frac{2x-2y+1}{\sqrt{2^2+(-2{)}^2}}$ and $Y=\frac{x+y-2}{\sqrt{1+1}}$

$\therefore$ The equation becomes $2Y^2=2\sqrt{2}X$
$\Rightarrow Y^2=\sqrt{2}x$.

$\therefore$ $4a=\sqrt{2}$
$a=\frac{\sqrt{2}}{4}$

The focus is the point of intersection of $X=a$ and $Y=0$
i.e. $\frac{2x-2y+1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$

i.e. $2x-2y=0$ and $x+y-2=0$

i.e. $S\equiv (1,1)$

Hence, option B.