Question

If $a{x}^3+b{y}^3+c{x}^{2}y+dx{y}^2=0$ represents three distinct straight lines such that each line bisects the angle between the other two, then which of the following is (are) CORRECT?

• A. $3b+c=0$
• B. $3c+b=0$
• C. $d^2>3bc$
• D. $d^2<3bc$

Answer 1

Answer: (A), (C)

The correct options are
A $3b+c=0$
C $d^2>3bc$

$a{x}^3+b{y}^3+c{x}^{2}y+dx{y}^2=0$
As the equation is homogeneous equation, so all the lines pass through the origin.
Let the equation of line be $y=mx$
Now, $a+b{m}^3+cm+d{m}^2=0$
$\Rightarrow b{m}^3+d{m}^2+cm+a=0\cdots \left(1\right)$
Equation $\left(1\right)$ should have three distinct real solutions.
Let $m_1,m_2,m_3$ be the solutions.

Every pair of lines makes same angle, assuming that angle to be $\theta$.
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\Rightarrow m_1-m_2=(1+m_1{m}_2)\tan \theta \cdots \left(2\right)$

Similarly, $m_2-m_3=(1+m_2{m}_3)\tan \theta \cdots \left(3\right)$
and $m_3-m_1=(1+m_3{m}_1)\tan \theta \cdots \left(4\right)$
Adding equation $\left(2\right),\left(3\right)$ and $\left(4\right)$, we get
$0=\tan \theta (3+m_1{m}_2+m_2{m}_3+m_3{m}_1)\Rightarrow (3+m_1{m}_2+m_2{m}_3+m_3{m}_1)=0[\because \tan \theta \ne 0]\Rightarrow m_1{m}_2+m_2{m}_3+m_3{m}_1=-3$
From equation $\left(1\right)$,
$\frac{c}{b}=-3\Rightarrow 3b+c=0$

As the equation $\left(1\right)$ has three distinct real roots, so it has two extrema.
Differentiating w.r.t. $m$,
$3b{m}^2+2dm+c=0$
It has two real and distinct roots.
So, $D>0$
$\Rightarrow 4d^2-12bc>0\Rightarrow d^2>3bc$