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If the slope of one of the lines represented by $a x^2-6 x y+$ $y^2=0$ is the square of the other, then the value of $a$ is
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The correct answer is:
$-27$ 0r $8$
Given pair of lines, $a x^2-6 x y+y^2=0$
Let the slope of one line be $m$, then slope of another line will be $m^2$.
We know that
On cubing Eq. (i) both sides, we get
$\begin{aligned}
& \left(m+m^2\right)^3=(6)^3 \\
& \Rightarrow \quad m^3+m^6+3 m^3\left(m+m^2\right)=216 \\
& \Rightarrow \quad m^3+m^6+18 m^3=216 \quad\left[\because m^3=a\right]
\end{aligned}$
$\begin{aligned} & \Rightarrow \quad a+a^2+18 a=216 \\ & \Rightarrow \quad a^2+19 a-216=0 \\ & a^2+27 a-8 a-216=0 \\ & a(a+27)-8(a+27)=0 \\ & (a+27)(a-8)=0 \\ & \therefore \quad a=-27,8\end{aligned}$
Let the slope of one line be $m$, then slope of another line will be $m^2$.
We know that
On cubing Eq. (i) both sides, we get
$\begin{aligned}
& \left(m+m^2\right)^3=(6)^3 \\
& \Rightarrow \quad m^3+m^6+3 m^3\left(m+m^2\right)=216 \\
& \Rightarrow \quad m^3+m^6+18 m^3=216 \quad\left[\because m^3=a\right]
\end{aligned}$
$\begin{aligned} & \Rightarrow \quad a+a^2+18 a=216 \\ & \Rightarrow \quad a^2+19 a-216=0 \\ & a^2+27 a-8 a-216=0 \\ & a(a+27)-8(a+27)=0 \\ & (a+27)(a-8)=0 \\ & \therefore \quad a=-27,8\end{aligned}$
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