Question

Let $m_1,m_2,m_3$ be the slopes of all the three straight lines represented by an equation $y^3+(2a+5)x{y}^2-6x^{2}y-2a{x}^3=0.$ If $a,m_1,m_2,m_3$ are all integers, then which of the following holds good?

• A. $a+\sum _{i=1}^3{m}_i=-1$
• B. $a+\sum _{i=1}^3{m}_i=-5$
• C. $a+\prod _{i=1}^3{m}_i=0$
• D. $a+\prod _{i=1}^3{m}_i=4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a+\sum _{i=1}^3{m}_i=-1$
B $a+\sum _{i=1}^3{m}_i=-5$
C $a+\prod _{i=1}^3{m}_i=0$

Given equation is $y^3+(2a+5)x{y}^2-6x^{2}y-2a{x}^3=0$
$\Rightarrow {\left(\frac{y}{x}\right)}^3+(2a+5){\left(\frac{y}{x}\right)}^2-6\left(\frac{y}{x}\right)-2a=0$
Substitute $\frac{y}{x}=m$
$m^3+(2a+5)m^2-6m-2a=0$
The roots of above equation are $m_1,m_2,m_3.$
$\therefore m_1+m_2+m_3=-(2a+5)$
$m_1{m}_2+m_2{m}_3+m_3{m}_1=-6$
$m_1{m}_2{m}_3=2a$

For $a+\sum _{i=1}^3{m}_i=-1$
$\Rightarrow a-(2a+5)=-1$
$\Rightarrow a=-4$ which is an integer.
$\Rightarrow m_1{m}_2{m}_3=-8$
$\Rightarrow m_1=1,m_2=-2,m_3=4$
So, here condition is satisfied.

For $a+\sum _{i=1}^3{m}_i=-5$
$\Rightarrow a-(2a+5)=-5$
$\Rightarrow a=0$ which is an integer.
$\Rightarrow m_1{m}_2{m}_3=0$
$\Rightarrow m_1=1,m_2=0,m_3=-6$
So, here also condition is satisfied.

For $a+\prod _{i=1}^3{m}_i=0$
$\Rightarrow a+2a=0$
$\Rightarrow a=0$ which is an integer
So, here also condition is satisfied.

For $a+\prod _{i=1}^3{m}_i=4$
$\Rightarrow a+2a=4$
$\Rightarrow a=\frac{4}{3}$ which is not an integer
So, here condition is not satisfied.