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Q. The distance between the pair of parallel lines $ x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ is :

BITSATBITSAT 2006

Solution:

Given equation is
$x^{2}+y^{2}+2 x y-8 a x-8 a y-9 a^{2}=0$
or $x^{2}+y^{2}+(-4 a)^{2}+2 x y-8 a x-8 a y-25 a^{2}=0$
or $(x+y-4 a)^{2}-(5 a)^{2}=0$
or $(x+y-9 a)(x +y +a)=0$
$\Rightarrow x+y-9 a=0$
or $x+y+a=0$
These lines are parallel. Now, we find the distance from origin to the line.
Let, $p_{1}=\frac{0+0-9 a}{\sqrt{1^{2}+1^{2}}}, p_{2}=\frac{0+0+a}{\sqrt{1^{2}+1^{2}}}$
$p_{1}=-\frac{9 a}{\sqrt{2}}, p_{2}=\frac{a}{\sqrt{2}}$
The distance between two lines is
$\left |p_{2}-p_{1}\right|=\left|\frac{a}{\sqrt{2}}+\frac{9 a}{\sqrt{2}}\right|$
$=\frac{10 a}{\sqrt{2}}$
$=5 \sqrt{2} a$