JEE Advanced 2025 Paper 2 · Q06 · Parabola

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 $R$ 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?

SolveKar AI · Accepted Answer

For the parabola $y^2 = x$ with chord midpoint $(h, k)$, the chord meets the parabola at $(y_1^2, y_1)$ and $(y_2^2, y_2)$ with $y_1 + y_2 = 2k$. The area between parabola and chord equals $\frac{1}{6}|y_2 - y_1|^3$. Now $y_2 - y_1 = 2\sqrt{h - k^2}$ (using the chord equation), so area $= \frac{4}{3}(h - k^2)^{3/2}$. Set $= \frac{4}{3}$ $	o$ $h - k^2 = 1$, i.e., the locus $S$ is $x = y^2 + 1$. (A) $(4, \sqrt{3})$: $4 - 3 = 1$. TRUE. (B) $(5, \sqrt{2})$: $5 - 2 = 3 
e 1$. FALSE. Region $R$ between $y = \sqrt{x}$ (parabola) and $y = \sqrt{x - 1}$ (curve $S$) in first quadrant, $x \in [1, 4]$: Area $= \int_1^4 [\sqrt{x} - \sqrt{x-1}]\, dx = \left[\frac{2}{3} x^{3/2} - \frac{2}{3}(x-1)^{3/2}ight]_1^4 = \frac{2}{3}(8 - 3\sqrt{3}) - \frac{2}{3}(1 - 0) = \frac{2}{3}(7 - 3\sqrt{3}) = \frac{14}{3} - 2\sqrt{3}$. (C) TRUE. Correct: (A), (C).

JEE Advanced 2025 Paper 2 · Q06 · Parabola

Solution