Question

If the distance between the parallel lines given by $x^2+2xy+y^2-9a^2=0$ is $90\sqrt{2}$ then $a$ is equal to

Answer 1

Let the equation of parallel lines be $ax+by+c_1=0andax+by+c_2=0$

Combined equation is $(ax+by+c_1)(ax+by+c_2)=0$

$a^2{x}^2+b^2{y}^2+2abxy+(ax+by)(c_1+c_2)+c_1{c}_2=0$

Comparing with $x^2+y^2+2xy-9a^2=0$

$a=1$ and $b=1$and $c_1=-c_2=c$because there are nox,y terms and $c_1{c}_2=-9a^2=>c^2=9a^2=>c=3a$

Distance between parallel lines $=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$
$=\frac{|2c|}{\sqrt{2}}$

Given distance $90\sqrt{2}$
$\therefore 90\sqrt{2}=\frac{|2c|}{\sqrt{2}}$

$\therefore |c|=90$

$\therefore |3a|=90$

$=>a=30or-30$