The point of intersection of the lines $\left(a^3+3\right) x+a y+a-3=0$ and $\left(a^5+2\right) x+(a+2) y+2…
The point of intersection of the lines $\left(a^3+3\right) x+a y+a-3=0$ and $\left(a^5+2\right) x+(a+2) y+2 a+3=0$ (a real) lies on the $y$-axis for
no value of $a$
more than two values of $a$
exactly one value of $a$
exactly two values of $a$
Solution
Given equation of lines are $\left(a^3+3\right) x+a y+a-3=0$ and $\left(a^5+2\right) x+(a+2) y+2 a+3=0$ (a real)
Since point of intersection of lines lies on $\mathrm{y}$-axis.
$\therefore$ Put $x=0$ in each equation, we get $a \mathrm{y}+a-3=0$ and $(a+2) y+2 a+3=0$
On solving these we get
$
\begin{aligned}
& (a+2)(a-3)-a(2 a+3)=0 \\
& \Rightarrow a^2-a-6-2 a^2-3 a=0 \\
& \Rightarrow \quad-a^2-4 a-6=0 \Rightarrow a^2+4 a+6=0 \\
& \Rightarrow \quad a=\frac{-4 \pm \sqrt{16-24}}{2}=\frac{-4 \pm \sqrt{-8}}{2} \\
& (\text { not real) }
\end{aligned}
$
This shows that the point of intersection of the lines lies on the $y$-axis for no value of ' $a$ '