Question

Let $S$ denote the locus of the mid-points of those chords of the parabola $y^2 = x$, such that the area of the region enclosed between the parabola and the chord is $\frac{4}{3}$. Let $\mathcal{R}$ denote the region lying in the first quadrant, enclosed by the parabola $y^2 = x$, the curve $S$, and the lines $x = 1$ and $x = 4$.

Then which of the following statements is (are) TRUE?

• A. $(4, \sqrt{3}) \in S$
• B. $(5, \sqrt{2}) \in S$
• C. Area of $\mathcal{R}$ is $\frac{14}{3} - 2\sqrt{3}$
• D. Area of $\mathcal{R}$ is $\frac{14}{3} - \sqrt{3}$

Answer 1

Answer: (A), (C)

(A), (C)

Answer 2

Let the parabola be $y^2 = x$. Let $M(h, k)$ be the midpoint of a chord of the parabola. The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$, where $S = y^2 - x$.

$yk - \frac{1}{2}(x + h) = k^2 - h$ $2yk - x - h = 2k^2 - 2h$ $x = 2yk - 2k^2 + h$.

The area of the region enclosed between the parabola $y^2 = x$ and the chord $x = 2yk - 2k^2 + h$ is given by integrating with respect to $y$. The intersection points of the parabola and the chord are found by substituting $x = y^2$ into the chord equation:

$y^2 = 2yk - 2k^2 + h$ $y^2 - 2ky + (2k^2 - h) = 0$.

Let the roots of this quadratic equation in $y$ be $y_1$ and $y_2$. These are the y-coordinates of the intersection points. From Vieta's formulas, $y_1 + y_2 = 2k$ and $y_1 y_2 = 2k^2 - h$. The area between the parabola $x = y^2$ and the chord is given by the integral $\int_{y_1}^{y_2} (x_{line} - x_{parabola}) dy$.

Area $= \int_{y_1}^{y_2} (2yk - 2k^2 + h - y^2) dy$. The integrand is $-(y^2 - 2ky + 2k^2 - h)$, which can be written as $-(y - y_1)(y - y_2)$. The value of the integral $\int_{y_1}^{y_2} -(y - y_1)(y - y_2) dy$ is $\frac{1}{6}(y_2 - y_1)^3$. We are given that this area is $\frac{4}{3}$.

$\frac{1}{6}(y_2 - y_1)^3 = \frac{4}{3}$ $(y_2 - y_1)^3 = 6 \times \frac{4}{3} = 8$ $|y_2 - y_1| = 2$. Assuming $y_2 > y_1$, we have $y_2 - y_1 = 2$.

We have the system of equations: $y_1 + y_2 = 2k$ $y_2 - y_1 = 2$

Adding the two equations gives $2y_2 = 2k + 2$, so $y_2 = k + 1$. Subtracting the second equation from the first gives $2y_1 = 2k - 2$, so $y_1 = k - 1$.

Now use the product of roots: $y_1 y_2 = 2k^2 - h$. $(k - 1)(k + 1) = 2k^2 - h$ $k^2 - 1 = 2k^2 - h$ $h = 2k^2 - k^2 + 1 = k^2 + 1$. Let $(h, k) = (x, y)$. The locus of the midpoints $S$ is given by the equation $x = y^2 + 1$.

Let's check the given statements about $S$: (A) $(4, \sqrt{3}) \in S$? Substitute $x=4, y=\sqrt{3}$ into $x = y^2 + 1$: $4 = (\sqrt{3})^2 + 1 = 3 + 1 = 4$. This is true. So $(4, \sqrt{3}) \in S$. (B) $(5, \sqrt{2}) \in S$? Substitute $x=5, y=\sqrt{2}$ into $x = y^2 + 1$: $5 = (\sqrt{2})^2 + 1 = 2 + 1 = 3$. This is false. So $(5, \sqrt{2}) \notin S$.

Now consider the region $\mathcal{R}$. $\mathcal{R}$ is in the first quadrant, enclosed by the parabola $y^2 = x$ (or $x = y^2$), the curve $S$ ($x = y^2 + 1$), and the lines $x = 1$ and $x = 4$.

The region $\mathcal{R}$ is the set of points $(x, y)$ in the first quadrant such that $1 \le x \le 4$ and $y^2 \le x \le y^2+1$. This means $y^2 \le x$ and $x \le y^2+1$. Combining with $1 \le x \le 4$: $1 \le x \le \min(4, y^2+1)$ and $\max(1, y^2) \le x \le 4$. The condition $y^2 \le x \le y^2+1$ means the region is between the two parabolas. The condition $1 \le x \le 4$ means the region is between the vertical lines.

Let's integrate with respect to $x$. Area $\mathcal{R} = \int_1^4 (y_{upper} - y_{lower}) dx$.

From $y^2 \le x$, we get $y \le \sqrt{x}$ (since $y \ge 0$). From $x \le y^2+1$, we get $y^2 \ge x-1$, so $y \ge \sqrt{x-1}$ (since $y \ge 0$). So the region is $1 \le x \le 4$ and $\sqrt{x-1} \le y \le \sqrt{x}$. Area $\mathcal{R} = \int_1^4 (\sqrt{x} - \sqrt{x-1}) dx$. Area $\mathcal{R} = \int_1^4 x^{1/2} dx - \int_1^4 (x-1)^{1/2} dx$. $\int x^{1/2} dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}$. $\int (x-1)^{1/2} dx = \frac{(x-1)^{3/2}}{3/2} = \frac{2}{3} (x-1)^{3/2}$. Area $\mathcal{R} = \left[ \frac{2}{3} x^{3/2} \right]_1^4 - \left[ \frac{2}{3} (x-1)^{3/2} \right]_1^4$. Area $\mathcal{R} = \left( \frac{2}{3} (4)^{3/2} - \frac{2}{3} (1)^{3/2} \right) - \left( \frac{2}{3} (4-1)^{3/2} - \frac{2}{3} (1-1)^{3/2} \right)$. Area $\mathcal{R} = \left( \frac{2}{3} (8) - \frac{2}{3} (1) \right) - \left( \frac{2}{3} (3)^{3/2} - \frac{2}{3} (0)^{3/2} \right)$. Area $\mathcal{R} = \left( \frac{16}{3} - \frac{2}{3} \right) - \left( \frac{2}{3} (3\sqrt{3}) - 0 \right)$. Area $\mathcal{R} = \frac{14}{3} - 2\sqrt{3}$.

Let's check the statements about the area of $\mathcal{R}$: (C) Area of $\mathcal{R}$ is $\frac{14}{3} - 2\sqrt{3}$. This is TRUE. (D) Area of $\mathcal{R}$ is $\frac{14}{3} - \sqrt{3}$. This is FALSE.

The correct statements are (A) and (C).