JEE Advanced 2025 · Paper 2 · Mathematics Q6

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MathematicsMultiple correct+4 / -2Question 6 of 48

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 $\dfrac43$ . 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?

(4, \sqrt3) \in S

(5, \sqrt2) \in S

Area of

\mathcal{R}

\dfrac{14}{3} - 2\sqrt3

Area of

\mathcal{R}

\dfrac{14}{3} - \sqrt3

Answer and worked solution

Official answer: A, C

Chord condition. For $x=y^2$ , the chord between ordinates $y_1,y_2$ cuts off area $\frac{|y_1-y_2|^3}{6}$ . Setting this to $\frac43$ gives $|y_1-y_2|=2$ .

Locus. With $y_{1,2}=k\pm1$ , the midpoint is
$\left(\frac{y_1^2+y_2^2}{2},\,k\right)=(k^2+1,\,k),$
so $S$ is $x=y^2+1$ .

(A) $(4,\sqrt3)$ : $3+1=4$ . ✓
(B) $(5,\sqrt2)$ : $2+1=3\ne5$ . ✗

Area of $\mathcal R$ .
$\int_1^4\left(\sqrt x-\sqrt{x-1}\right)dx=\frac23(8-1)-\frac23\cdot3\sqrt3=\frac{14}3-2\sqrt3.$
So (C) is true and (D) is false.

Topics:Area between curves Parabola Locus

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JEE Advanced 2025 · Paper 2, Mathematics question 6