Question

Shew that two of the straight lines represented by the equation $a{y}^4+bx{y}^3+c{x}^2{y}^2+d{x}^{3}y+w{x}^4=0$ will be at right angles if $(b+d)(ad+be)+(e-a{)}^2(a+c+e)=0$.

Answer 1

The equation is $a{y}^4+bx{y}^3+c{x}^2{y}^2+d{x}^{3}y+e{x}^4=0$
If two of the lines represented by $1$ be at right angles, their equation must be of the form $x^2+pxy-y^2=0$ as the sum of the co efficients of $x^2$ & $y^2$ is zero
Hence let,
$a{y}^4+bx{y}^3+c{x}^2{y}^2+d{x}^{3}y+e{x}^4=(x^2+pxy-y^2)({ex}^2+qxy-a{y}^2)$
Equating co efficients
$-pa-q=b....\left(2\right)$
$\left(x^2{y}^2\right);-a+pq-e=c....\left(3\right)$
$\left(x^{2}y\right);q+ep=d....\left(4\right)$
From $2$ and $4$
$p=\frac{b+d}{e-a}$ and $q=\frac{-da-eb}{e-a}$
Substituting the values in $3$ we get
$-a+\frac{b+d}{e-a}(-\frac{da+eb}{e-a})-e=c$
or $(a+e+c){(e-a)}^2+(b+d)(ad+be)=0$