Question

The equation of second degree $x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0$ represents a pair of straight lines, the distance between them is?

• A. $3$
• B. $2\sqrt{3}$
• C. $4$
• D. $4\sqrt{3}$

Answer 1

The correct option is A $3$

The equation is $x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0$

Here, $a=1,b=2,h=\sqrt{2}$

So, $h^2=ab$ which means it represents a pair of parallel lines.

$x^2+2\sqrt{2}xy+(\sqrt{2}y{)}^2+4(x+\sqrt{2}y)+1=0$

$\Rightarrow (x+\sqrt{2}y{)}^2+4(x+\sqrt{2}y)+1=0$

Let $x+\sqrt{2}y=m$

$\Rightarrow m^2+4m+1=0\Rightarrow m=\frac{-4\pm \sqrt{16-4}}{2}$

$=-2\pm \sqrt{3}$

So, the lines are $x+\sqrt{2}y+2-\sqrt{3}=0$ and $x+\sqrt{2}y+2+\sqrt{3}=0$

$\therefore$ Distance between them = $\frac{2+\sqrt{3}-2+\sqrt{3}}{\sqrt{1+2}}$

$=\frac{2\sqrt{3}}{\sqrt{3}}$

$=3$ units