Question

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

• A. $2\sqrt{5}a$
• B. $\sqrt{10}a$
• C. $10a$
• D. $5\sqrt{2}a$

Answer 1

Answer: (D)

Given equation is
$x^2+y^2+2xy-8ax-8ay-9a^2=0$
or $x^2+y^2+(-4a{)}^2+2xy-8ax-8ay-25a^2=0$
or $(x+y-4a{)}^2-(5a{)}^2=0$
or $(x+y-9a)(x+y+a)=0$
$\Rightarrow x+y-9a=0$
or $x+y+a=0$
These lines are parallel. Now, we find the distance from origin to the line.
Let, $p_1=\frac{0+0-9a}{\sqrt{1^2+1^2}},p_2=\frac{0+0+a}{\sqrt{1^2+1^2}}$
$p_1=-\frac{9a}{\sqrt{2}},p_2=\frac{a}{\sqrt{2}}$
The distance between two lines is
$\left|p_2-p_1\right|=\left|\frac{a}{\sqrt{2}}+\frac{9a}{\sqrt{2}}\right|$
$=\frac{10a}{\sqrt{2}}$
$=5\sqrt{2}a$