The point $(a^2, a+1)$ lies in the angle between the lines $3x - y + 1 = 0$ and $x + 2y - 5 = 0$ containing the origin then the possible integral value of $a$ is/are:
Step-by-Step Solution
Key Concept: Finding line-parabola intersections requires substituting the line equation into the parabola equation and solving the resulting quadratic or linear equations.
To find the intersection of the parabola $x = (y-1)^2$ with lines, solve $(y-1)^2 = \frac{y-1}{3}$ to get $y=1$ or $y=\frac{4}{3}$, yielding points $P(\frac{1}{9}, \frac{4}{3})$. For the second condition $(y-1)^2 = 5-2y$, rearrange to $y^2 = 4$, so $y = \pm 2$, giving points $Q(1,2)$ and $R(9,-2)$. The parameter $a$ lies in $(-3, 0) \cup (\frac{1}{3}, 1)$ for the required geometric configuration.
Correct Answer: 2,4