Question

Prove that the equation $8x^2+8xy+2y^2+26x+13y+15=0$, represents two parallel straight lines and find the distance between them.

Answer 1

Here $h^2-ab=(4{)}^2-2\times 8=0$.
$\therefore$ the lines are parallel.
On dividing by $2$,
$4x^2+4xy+y^2+13x+\frac{13}{2}y+\frac{15}{2}=0$
Suppose its factors are $(2x+y+p)(2x+y+q)$
On comparing the coefficients, we get
$2(p+q)=13,p+q=\frac{13}{2}$ and $pq=\frac{15}{2}$,
or $p(\frac{13}{2}-p)=\frac{15}{2}$ or $2p^2-13p+15=0$
$\therefore (p-5)(2p-3)=0$
or $p=5,\frac{3}{2}$;
$\therefore q=\frac{3}{2},5$
$\therefore 2x+y+5=0$, $2x+y+\frac{3}{2}=0$
are the required lines.
If $p_1$ and $p_2$ be their distances from origin, then the distance between them is
$p=p_1\sim p_2=\frac{5}{\sqrt{(4+1)}}\sim \frac{3/2}{\sqrt{(4+)}}=\frac{7}{2\sqrt{5}}$.