Question

Prove that two of the lines represented by the equation
$a{y}^4+bx{y}^3+c{x}^2{y}^2+d{x}^{3}y+e{x}^4=0$
will be perpendicular if
$(b+d)(ad+be)+(e-a{)}^2(a+c+e)=0$

Answer 1

Arguing as in $Q.5$ above one of the factors of the given
equation will be $x^2+pxy-y^2=0$.
Now keeping in view the coefficients of $x^4$ and $y^4$ the other factor will be
$e{x}^2+qxy-a{y}^2=0$
$\therefore \quad ay^{ 4 }+bxy^{ 3 }+cx^{ 2 }y^{ 2 }+dx^{
3 }y+ex^{ 4 }$
$=(x^2+pxy-y^2)(e{x}^2+qxy-a{y}^2)$.
Comparing the coefficients,
$b=-ap-q\left(byx{y}^3\right)$
$d=ep+q\left(by{x}^{3}y\right)$
$c=-a-e+pq\left(by{x}^2{y}^2\right)$
Solving $\left(1\right)$ and $\left(2\right)$ for $p$ and $q$,
$p=\frac{b+d}{e-d}$ and $q=-\dfrac{ ad+be }{ e-a
}$.
On putting the values of $p$ and $q$ in $\left(3\right)$, we get
$(c+a+e)=-\frac{(b+d)(ad+be)}{(e-a{)}^2}$
or $(a+c+e)(e-a{)}^2+(b+d\left)\right(ad+be)=0$.