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MathHardAdv 2025 • Paper 2

Locus of Midpoints of Parabola Chords and Enclosed Area

Let SS denote the locus of the mid-points of those chords of the parabola y2=xy^2 = x, such that the area of the region enclosed between the parabola and the chord is 43\frac{4}{3}. Let R\mathcal{R} denote the region lying in the first quadrant, enclosed by the parabola y2=xy^2 = x, the curve SS, and the lines x=1x = 1 and x=4x = 4.

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

Options

A

(4,3)∈S(4, \sqrt{3}) \in S

Correct
B

(5,2)∈S(5, \sqrt{2}) \in S

C

Area of R\mathcal{R} is 143−23\frac{14}{3} - 2\sqrt{3}

Correct
D

Area of R\mathcal{R} is 143−3\frac{14}{3} - \sqrt{3}

Topics & Concepts

Conic SectionsParabolaArea Under Curves

Step-by-Step Solution

To determine the correct statements, we first derive the equation of the locus SS of the mid-points of the chords.

Let the endpoints of a chord of the parabola y2=xy^2 = x be P(y12,y1)P(y_1^2, y_1) and Q(y22,y2)Q(y_2^2, y_2).

The area of the region enclosed between the parabola y2=xy^2 = x and the chord PQPQ is given by the standard formula: Area=∣y1−y2∣36\text{Area} = \frac{|y_1 - y_2|^3}{6}

Given that this area is 43\frac{4}{3}, we have: ∣y1−y2∣36=43  ⟹  ∣y1−y2∣3=8  ⟹  ∣y1−y2∣=2\frac{|y_1 - y_2|^3}{6} = \frac{4}{3} \implies |y_1 - y_2|^3 = 8 \implies |y_1 - y_2| = 2

Let (h,k)(h, k) be the mid-point of the chord PQPQ. Then: k=y1+y22  ⟹  y1+y2=2kk = \frac{y_1 + y_2}{2} \implies y_1 + y_2 = 2k h=x1+x22=y12+y222h = \frac{x_1 + x_2}{2} = \frac{y_1^2 + y_2^2}{2}

Using the algebraic identity (y1−y2)2=(y1+y2)2−4y1y2(y_1 - y_2)^2 = (y_1 + y_2)^2 - 4y_1 y_2, we substitute ∣y1−y2∣=2|y_1 - y_2| = 2 and y1+y2=2ky_1 + y_2 = 2k: 22=(2k)2−4y1y2  ⟹  4=4k2−4y1y2  ⟹  y1y2=k2−12^2 = (2k)^2 - 4y_1 y_2 \implies 4 = 4k^2 - 4y_1 y_2 \implies y_1 y_2 = k^2 - 1

Now, expressing hh in terms of kk: h=(y1+y2)2−2y1y22=4k2−2(k2−1)2=k2+1h = \frac{(y_1 + y_2)^2 - 2y_1 y_2}{2} = \frac{4k^2 - 2(k^2 - 1)}{2} = k^2 + 1

Replacing (h,k)(h, k) with (x,y)(x, y), the equation of the locus SS is: x=y2+1  ⟹  y2=x−1x = y^2 + 1 \implies y^2 = x - 1

Step 1: Verification of Options (A) and (B)

  • For Option (A): Point (4,3)(4, \sqrt{3}) (3)2=3and4−1=3(\sqrt{3})^2 = 3 \quad \text{and} \quad 4 - 1 = 3 Since 3=33 = 3, (4,3)∈S(4, \sqrt{3}) \in S. Thus, Option (A) is TRUE.

  • For Option (B): Point (5,2)(5, \sqrt{2}) (2)2=2and5−1=4≠2(\sqrt{2})^2 = 2 \quad \text{and} \quad 5 - 1 = 4 \neq 2 Thus, Option (B) is FALSE.

Step 2: Calculation of the Area of Region R\mathcal{R}

The region R\mathcal{R} lies in the first quadrant and is bounded above by y=xy = \sqrt{x} (from y2=xy^2 = x), below by y=x−1y = \sqrt{x-1} (from the locus SS), and by the vertical lines x=1x = 1 and x=4x = 4.

The area of R\mathcal{R} is given by: Area(R)=∫14(x−x−1)dx\text{Area}(\mathcal{R}) = \int_{1}^{4} \left( \sqrt{x} - \sqrt{x-1} \right) dx

Evaluating the integral step-by-step: ∫14x dx=[23x3/2]14=23(43/2−13/2)=23(8−1)=143\int_{1}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4} = \frac{2}{3} \left( 4^{3/2} - 1^{3/2} \right) = \frac{2}{3} (8 - 1) = \frac{14}{3}

∫14x−1 dx=[23(x−1)3/2]14=23(33/2−0)=23(33)=23\int_{1}^{4} \sqrt{x-1} \, dx = \left[ \frac{2}{3} (x-1)^{3/2} \right]_{1}^{4} = \frac{2}{3} \left( 3^{3/2} - 0 \right) = \frac{2}{3} (3\sqrt{3}) = 2\sqrt{3}

Subtracting the two integrals: Area(R)=143−23\text{Area}(\mathcal{R}) = \frac{14}{3} - 2\sqrt{3}

Thus, Option (C) is TRUE and Option (D) is FALSE.

Conclusion

The correct statements are A and C.

Locus of Midpoints of Parabola Chords and Enclosed Area | Mathematics PYQ Solution - JEE Challenger