CUET UG 2025 — Mathematics Calculus

Given parabola: $y^2 = 8x$y2=8x

Comparing with the standard form $y^2 = 4ax$y2=4ax:

$$4a = 8$$4a=8

$$a = 2$$a=2

The latus rectum is the vertical line $x = a = 2$x=a=2

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To find where the latus rectum meets the parabola, substitute $x = 2$x=2:

$$y^2 = 8(2) = 16$$y2=8(2)=16

$y = 4$y=4 (in the first quadrant)

The region is bounded by the parabola from $x = 0$x=0 to $x = 2$x=2, the latus rectum at $x = 2$x=2, and the x-axis.

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From $y^2 = 8x$y2=8x:

$y = \sqrt{8x} = 2\sqrt{2x}$y=8x​=22x​ (positive in first quadrant)

Area $\displaystyle = \int_0^2 y \, dx$=∫02​ydx

$$= \int_0^2 2\sqrt{2x} \, dx$$=∫02​22x​dx

$$= 2\sqrt{2} \int_0^2 \sqrt{x} \, dx$$=22​∫02​x​dx

$$= 2\sqrt{2} \cdot \left[\frac{x^{3/2}}{3/2}\right]_0^2$$=22​⋅[3/2x3/2​]02​

$$= 2\sqrt{2} \cdot \frac{2}{3} \cdot [x^{3/2}]_0^2$$=22​⋅32​⋅[x3/2]02​

$$= \frac{4\sqrt{2}}{3} \cdot (2^{3/2} - 0)$$=342​​⋅(23/2−0)

$$= \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2}$$=342​​⋅22​

$$= \frac{8 \cdot 2}{3}$$=38⋅2​

$= \frac{16}{3}$=316​ square units