Question

If two of the lines represented by $x^4+x^{3}y+c{x}^2{y}^2-x{y}^3+y^4=0$ bisect the angle between the other two, then the value of $c$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $-6$

Answer 1

Answer: (D)

$-6$

Since the product of the slopes of the four lines represented by the given equation is $1$ and a pair of lines represent the bisectors of the angles between the other two, the product of the slopes of each pair is $-1$. So let the equation of one pair be $a{x}^2+2hxy-a{y}^2=0$

The equation of its bisectors is $\frac{x^2-y^2}{2a}=\frac{xy}{h}$

By hypothesis $x^4+x^{3}y+c{x}^2{y}^2-x{y}^3+y^4$ $=(a{x}^2+2hxy-a{y}^2)(h{x}^2-2axy-h{y}^2)$

$=ah(x^4+y^4)+2(h^2-a^2)(x^{3}y-x{y}^3)-6ah{x}^2{y}^2$

Comparing the respective coefficients we get

$ah=1$ and $c=-6ah=-6$