Question

The point of intersection of the lines $(a^3+3)x+ay+a-3=0$ and $(a^5+2)x+(a+2)y+2a+3=0$ (a real) lies on the y-axis for

• A. no value of $a$
• B. more than two values of $a$
• C. exactly one value of $a$
• D. exactly two values of $a$

Answer 1

Answer: (B)

Given equation of lines are
$(a^3+3)x+ay+a-3=0$ and $(a^5+2)x+(a+2)y+2a+3=0$ (a real)
Since point of intersection of lines lies on y-axis.
$\therefore$ Put $x=0$ in each equation, we get
$fly+a-3=0$ and
$(a+2)y+2a+3=0$
On solving these we get
$(a+2)(a-3)-a(2a+3)=0$
$\Rightarrow a^2-a-6-2a^2-3a=0$
$\Rightarrow -a^2-a-6-2a^2-3a=0\Rightarrow a^2+4a+6=0$
$\Rightarrow a=\frac{-4\pm \sqrt{16-24}}{2}=\frac{-4\pm \sqrt{-8}}{2}$ (not real)
This shows that the point of intersection of the lines lies on the $y$-axis for no value of ${}^{\prime }{a}^{\prime }$