Question

The number of integral values of a for which the point $P(a^2, a)$ lies in the region corresponding to the acute angle between the lines 2y = x and 4y = x is

Answer 1

1

Answer 2

To determine the region corresponding to the acute angle between two lines $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$, we first check the sign of $A_1A_2 + B_1B_2$.

The given lines are: $L_1: x - 2y = 0$ $L_2: x - 4y = 0$

Here, $A_1 = 1$, $B_1 = -2$, $C_1 = 0$. And $A_2 = 1$, $B_2 = -4$, $C_2 = 0$.

Calculate $A_1A_2 + B_1B_2$: $A_1A_2 + B_1B_2 = (1)(1) + (-2)(-4) = 1 + 8 = 9$.

Since $A_1A_2 + B_1B_2 > 0$, the origin $(0,0)$ lies in the obtuse angle between the lines. Therefore, a point $(x,y)$ lies in the acute angle region if the product of the expressions for the lines, evaluated at that point, is negative: $(x - 2y)(x - 4y) < 0$.

The given point is $P(a^2, a)$. Substitute $x = a^2$ and $y = a$ into the inequality: $(a^2 - 2a)(a^2 - 4a) < 0$

Factor out 'a' from each term: $a(a - 2) \cdot a(a - 4) < 0$ $a^2 (a - 2)(a - 4) < 0$

We need to find the integral values of 'a' that satisfy this inequality.

Case 1: $a = 0$. If $a = 0$, the inequality becomes $0^2 (0 - 2)(0 - 4) = 0 \cdot (-2) \cdot (-4) = 0$. $0 < 0$ is false. So $a=0$ is not a solution.

Case 2: $a \neq 0$. If $a \neq 0$, then $a^2$ is always positive ($a^2 > 0$). We can divide the inequality by $a^2$ without changing the direction of the inequality sign: $(a - 2)(a - 4) < 0$

This inequality holds true when 'a' is strictly between the roots 2 and 4. So, $2 < a < 4$.

We are looking for integral values of 'a'. The only integer that satisfies $2 < a < 4$ is $a=3$.

Thus, there is only one integral value of 'a' for which the point $P(a^2, a)$ lies in the region corresponding to the acute angle between the lines.