Question

The equation $a{x}^3-9y{x}^2-y^{2}x+4y^3=0$ represents three straight lines. If two of the lines are perpendicular to each other, then the value of $a$ is:

• A. 5
• B. -5
• C. 4
• D. -4

Answer 1

Answer: (A), (D)

The correct options are
A 5
D -4

The given equation is $a{x}^3-9y{x}^2-y^{2}x+4y^3=0$, which represents three straight lines.

We can see that all the lines passes from origin $(0,0)$

Let's assume the lines are given by equation $y=mx$

Putting $y=mx$ in the given equation of three straight lines, we get,

$\Rightarrow a{x}^3-9\left(mx\right)\left(x^2\right)-(mx{)}^{2}x+4(mx{)}^3=0$

$\Rightarrow (a-9m-m^2+4m^3)x^3=0$

$\Rightarrow 4m^3-m^2-9m+a=0$ ....$\left(1\right)$

This equation in $m$ has three roots, $m_1$, $m_2$ and $m_3$, which are three slopes of three lines respectively.

If the two lines are perpendicular then let's assume $m_1.m_2=-1$,

Product of roots in equation $\left(1\right)$ is $m_1.m_2.m_3=\frac{-a}{4}$

Hence $m_3=\frac{a}{4}$

also from equation $\left(1\right)$, $m_1+m_2+m_3=\frac{1}{4}$

$\Rightarrow m_1{m}_2+m_3(m_1+m_2)=\frac{-9}{4}$

$\Rightarrow m_1+m_2=\frac{1-a}{4}$

$\Rightarrow a(1-a)=-20$

By Solving the above equation we get $a=5,-4$