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 $'a'$