Question

Find the value of k so that the following equation may represent pairs of straight lines:
$12{{x}^{2}}-kxy+2{{y}^{2}}+11x-5y+2=0$

Answer 1

Hint: Compare the given equation $12{{x}^{2}}-kxy+2{{y}^{2}}+11x-5y+2=0$ with a general equation in second degree $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ then apply the condition for a pair of straight lines.

Complete step-by-step answer:
The equation $12{{x}^{2}}-kxy+2{{y}^{2}}+11x-5y+2=0$ given in the question is an equation in second degree so we can compare this equation by $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
 $12{{x}^{2}}-kxy+2{{y}^{2}}+11x-5y+2=0$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
Comparing the coefficient of ${{x}^{2}}$ will give a = 12.
Comparing the coefficient of $xy$ will give $h=-\dfrac{k}{2}$.
Comparing the coefficient of ${{y}^{2}}$ will give b = 2.
Comparing the coefficient of $x$ will give $g=\dfrac{11}{2}$.
Comparing the coefficient of $y$ will give $f=-\dfrac{5}{2}$.
Comparing the constant will give $c$ = 2.
The given equation represents a pair of straight lines so the condition for a general second degree equation $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ is:
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{f}^{2}}=0$
Substituting the value of a, b, c, f, g, h in the above equation we get,
$\begin{align}
  & \left( 12 \right)\left( 2 \right)\left( 2 \right)+2\left( -\dfrac{5}{2} \right)\left( \dfrac{11}{2} \right)\left( -\dfrac{k}{2} \right)-12\left( \dfrac{25}{4} \right)-2\left( \dfrac{121}{4} \right)-2\left( \dfrac{25}{4} \right)=0 \\
 & \Rightarrow 48+\dfrac{55k}{4}-75-\dfrac{121}{2}-\dfrac{25}{2}=0 \\
 & \Rightarrow -27-\dfrac{146}{2}+\dfrac{55k}{4}=0 \\
 & \Rightarrow \dfrac{55k}{4}-27-73=0 \\
 & \Rightarrow \dfrac{55k}{4}-100=0 \\
 & \Rightarrow k=\dfrac{400}{55} \\
 & \Rightarrow k=\dfrac{80}{11} \\
\end{align}$
Hence, the value of k for which the given equation represents a pair of straight lines is$\dfrac{80}{11}$.

Note: In the above problem, you might be wondering from where this condition for straight lines has come. The derivation for condition of a second degree equation represents a pair of straight lines is as follows:
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
$a{{x}^{2}}+\left( 2hy+2g \right)x+b{{y}^{2}}+2fy+c=0$
The above equation is quadratic in x so the solution is:
$x=\dfrac{-\left( 2hy+2g \right)\pm \sqrt{{{\left( 2hy+2g \right)}^{2}}-4\left( a{{x}^{2}} \right)\left( b{{y}^{2}}+2fy+c \right)}}{2a}$
$x=\dfrac{-\left( hy+g \right)\pm \sqrt{{{\left( hy+g \right)}^{2}}-\left( a{{x}^{2}} \right)\left( b{{y}^{2}}+2fy+c \right)}}{a}$
For the equation to be pair of straight lines, expression in the square root must be a perfect square so:
\[\begin{align}
  & {{h}^{2}}{{y}^{2}}+{{g}^{2}}+2hyg-ab{{y}^{2}}-2afy-ac \\
 & \left( {{h}^{2}}-ab \right){{y}^{2}}+\left( 2gh-2af \right)y+\left( {{g}^{2}}-ac \right) \\
\end{align}\]
The above quadratic equation in y is a perfect square when the discriminant of the above equation is 0.
D = 0 for \[\left( {{h}^{2}}-ab \right){{y}^{2}}+\left( 2gh-2af \right)y+\left( {{g}^{2}}-ac \right)\]
$D={{\left( 2gh-2af \right)}^{2}}-4\left( {{h}^{2}}-ab \right)\left( {{g}^{2}}-ac \right)=0$
Solving the above equation will give:
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{f}^{2}}=0$