Question

Prove that two of the lines represented by the equation
$a{x}^4+b{x}^3+c{x}^2{y}^2+dx{y}^3+a{y}^4=0$
will bisect the angles between the other two if
c + 6a = 0 and b + d = 0.

Answer 1

Let one pair be $p{x}^2+2qxy+r{y}^2=\mathrm{0.....}\left(i\right)$

Equation of angle bisector

$\frac{x^2-y^2}{p-q}=\frac{xy}{q}=0x^2-y^2-\frac{p-q}{r}xy=\mathrm{0......}\left(ii\right)$

Lines by $\left(i\right)$ and $\left(ii\right)$ is given by

$a{x}^4+b{x}^{3}y+c{x}^2{y}^2+dx{y}^3+a{y}^4=0$

$a{x}^4+b{x}^{3}y+c{x}^2{y}^2+dx{y}^3+a{y}^4=(p{x}^2+2qxy+r{y}^2)(x^2-y^2-\frac{p-r}{q}xy)....\left(iii\right)$

Comparing the coefficients of $x^4$ and$y^4$

$\Rightarrow p=a,r=-a$

Substituting in $\left(iii\right)$

$a{x}^4+b{x}^{3}y+c{x}^2{y}^2+dx{y}^3+{ay}^4=(p{x}^2+2qxy+r{y}^2)(x^2-y^2-\frac{2a}{q}xy)$

$a{x}^4+b{x}^{3}y+c{x}^2{y}^2+d{xy}^3+{ay}^4=(a{x}^4+-a{x}^2{y}^2-\frac{2a^2}{q}{x}^{3}y+2qb{x}^{3}y-2q{xy}^3-4a{x}^2{y}^2-a{x}^2{y}^2+a{y}^4+\frac{2a^2}{q}x{y}^3=0a{x}^4+b{x}^{3}y+c{x}^2{y}^2+d{xy}^3+{ay}^4=(a{x}^4+(2q-\frac{2a^2}{q})x^{3}y-6a{x}^2{y}^2+(\frac{2a^2}{q}-2q)x{y}^3+a{y}^4)$

Comparing the coefficients we get

$b=2q-\frac{2a^2}{q}......\left(iv\right)c=-6a............\left(v\right)d=\frac{2a^2}{q}-2q........\left(vi\right)$

Adding $\left(iv\right)$ and $\left(vi\right)$

$\Rightarrow b+d=0$

From $\left(v\right)$
$c+6a=0$

Hence proved