Area Under Curves — JEE Advanced · 2025 · 18 May 2025 · Paper 2 Question with Solution

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?

Answer: (A), (C)

Solution

For a chord of

y^{2} = x

with midpoint

(h, k)

, the chord equation is

k y = \frac{x + h}{2}

, or

x - 2 k y + 2 k^{2} - h = 0

.

Interchanging

x

and

y

for area calculation:

y - 2 k x + (2 k^{2} - h) = 0

intersects

y = x^{2}

.

The area between the parabola and chord is

\frac{∣ α - β ∣ ^{3}}{6}

where

α, β

are roots.

From

x^{2} - 2 k x + (2 k^{2} - h) = 0

∣ α - β ∣ = 4 k^{2} - 4 (2 k^{2} - h) = 2 h - k^{2}

.

Area

= \frac{( 2 h - k ^{2} ) ^{3}}{6} = \frac{4 ( h - k ^{2} ) ^{3/2}}{3} = \frac{4}{3}

.

So

(h - k^{2})^{3/2} = 1

, giving

h - k^{2} = 1

, or

h = k^{2} + 1

.

Locus

S

x = y^{2} + 1

.

(A)

(4, 3)

4 = 3 + 1

✓
(B)

(5, 2)

5 \neq = 2 + 1

✗

Area of

R

\int_{1}^{4} (x - x - 1) d x = \frac{2}{3} [x^{3/2} - (x - 1)^{3/2}]_{1}^{4} = \frac{2}{3} (8 - 33 - 1) = \frac{14}{3} - 23

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · One or more correct options