Question

lf two 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)

$-6$

Let, $a{x}^2+2hxy+b{y}^2=0$ be a pair of straight lines
This has pair of angle bisectors given by
$x^2-y^2=\frac{(a-b)}{h}xy$
$h{x}^2-h{y}^2=(a-b)xy$
$h{x}^2-h{y}^2-(a-b)xy=0$
Multiply these two pairs of lines we have,
$(a{x}^2+2hxy+b{y}^2)(h{x}^2-h{y}^2-(a-b\left)xy\right)=0$
$ah{x}^4+2h^2{x}^{3}y+bh{x}^2{y}^2-ah{x}^2{y}^2-2h^{2}x{y}^3-bh{y}^4-a(a-b)x^{3}y-2(a-b)hxy-b(a-b)x{y}^3=0$
$ah{x}^4-bh{y}^4+(2h^2-a^2+ab)x^{3}y-(2h^2-b^2+ab)y^{3}x+(bh-ab-2ab+2bh)xy=0$
$ah{x}^4+(2h^2-a^2+ab)x^{3}y+(3bh-3ah)xy+(b^2-2h^2-ab)y^{3}x-bh{x}^4=0$
Comparing the given equation with this equation we have,
$ah=1$ and $bh=-1$
$2h^2-a^2+ab=1$ and $3h(b-a)=c$
$3(bh-ah)=c\Rightarrow 3(-1-1)=c$
$c=-6$