Question

If $(a^2, a - 2)$ be a point interior to the region of the parabola $y^2 = 2x$ bounded by the chord joining the points $(2, 2)$ and $(8, -4)$, then the number of all possible integral values of a is :

• A. 5
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

To determine the number of possible integral values of 'a' for which the point $(a^2, a-2)$ is interior to the region bounded by the parabola $y^2 = 2x$ and the chord joining the points $(2, 2)$ and $(8, -4)$, we need to satisfy two conditions:

1. The point must be interior to the parabola $y^2 = 2x$. For a parabola $y^2 = 4Ax$, a point $(x_1, y_1)$ is interior if $y_1^2 - 4Ax_1 < 0$. In this case, $y^2 = 2x$ (so $4A=2$). The point is $(a^2, a-2)$. Therefore, we must have: $(a-2)^2 - 2(a^2) < 0$ $a^2 - 4a + 4 - 2a^2 < 0$ $-a^2 - 4a + 4 < 0$ $a^2 + 4a - 4 > 0$

   To find the roots of $a^2 + 4a - 4 = 0$, we use the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $a = \frac{-4 \pm \sqrt{4^2 - 4(1)(-4)}}{2(1)}$ $a = \frac{-4 \pm \sqrt{16 + 16}}{2}$ $a = \frac{-4 \pm \sqrt{32}}{2}$ $a = \frac{-4 \pm 4\sqrt{2}}{2}$ $a = -2 \pm 2\sqrt{2}$

   Approximate values for the roots are: $a_1 = -2 - 2\sqrt{2} \approx -2 - 2(1.414) = -2 - 2.828 = -4.828$ $a_2 = -2 + 2\sqrt{2} \approx -2 + 2(1.414) = -2 + 2.828 = 0.828$

   Since the parabola $a^2 + 4a - 4$ opens upwards, the inequality $a^2 + 4a - 4 > 0$ holds when $a < -2 - 2\sqrt{2}$ or $a > -2 + 2\sqrt{2}$. So, $a \in (-\infty, -4.828) \cup (0.828, \infty)$.

2. The point must be on the correct side of the chord. First, find the equation of the chord joining $(2, 2)$ and $(8, -4)$. Using the two-point form $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$: $\frac{y - 2}{x - 2} = \frac{-4 - 2}{8 - 2}$ $\frac{y - 2}{x - 2} = \frac{-6}{6}$ $\frac{y - 2}{x - 2} = -1$ $y - 2 = -(x - 2)$ $y - 2 = -x + 2$ $x + y - 4 = 0$

   Let $L(x,y) = x + y - 4$. The region bounded by the parabola and the chord is a parabolic segment. The point $(a^2, a-2)$ must lie on the same side of the chord as the vertex of the parabola $(0,0)$. Substitute $(0,0)$ into $L(x,y)$: $L(0,0) = 0 + 0 - 4 = -4$. Since $L(0,0) < 0$, the point $(a^2, a-2)$ must also satisfy $L(a^2, a-2) < 0$. $a^2 + (a-2) - 4 < 0$ $a^2 + a - 6 < 0$

   Factor the quadratic expression: $(a+3)(a-2) < 0$

   The roots are $a = -3$ and $a = 2$. Since the parabola $a^2+a-6$ opens upwards, the inequality $a^2 + a - 6 < 0$ holds when $a$ is between the roots. So, $a \in (-3, 2)$.

Combining the conditions: We need to find the integral values of 'a' that satisfy both conditions:

1. $a \in (-\infty, -4.828) \cup (0.828, \infty)$
2. $a \in (-3, 2)$

Let's find the intersection of these two intervals: Intersection with $(-\infty, -4.828)$: The interval $(-3, 2)$ does not overlap with $(-\infty, -4.828)$, so this part yields an empty set. Intersection with $(0.828, \infty)$: The intersection of $(-3, 2)$ and $(0.828, \infty)$ is $(0.828, 2)$.

So, the combined condition for 'a' is $0.828 < a < 2$.

The integral values of 'a' that satisfy this condition are $a=1$. There is only one possible integral value of 'a'.