JEE Main 2024 — 30 January, Shift 1 — Question 8

Substitute $y = 2x - 8$ into the parabola: $\Rightarrow (2x - 8)^2 = 4(x - 2)$ $\Rightarrow 4x^2 - 32x + 64 = 4x - 8$ $\Rightarrow 4x^2 - 36x + 72 = 0$ $\Rightarrow x^2 - 9x + 18 = 0$ $\Rightarrow x = 3, 6$

Find corresponding $y: \quad$ $x = 3 \Rightarrow y = 2(3) - 8 = -2, \quad$ $x = 6 \Rightarrow y = 2(6) - 8 = 4$ $\Rightarrow$ Points of intersection: $(3, -2), (6, 4)$

Since $y^2 = 4(x - 2) \Rightarrow x = \frac{y^2}{4} + 2$ $\Rightarrow$ Area between curves from $y = -2$ to $y = 4:$ $A = \int_{-2}^{4} \left[ (2x - 8) - \left( -\sqrt{4(x - 2)} \right) \right] dx$ is hard to set up in $dx$ due to inverse

So use horizontal strip: $x_{\text{right}} = \frac{y^2}{4} + 2, \quad x_{\text{left}} = \frac{y + 8}{2}$

Area $A = \int_{-2}^{4} \left( \frac{y^2}{4} + 2 - \frac{y + 8}{2} \right) dy$ $= \int_{-2}^{4} \left( \frac{y^2}{4} + 2 - \frac{y}{2} - 4 \right) dy$ $= \int_{-2}^{4} \left( \frac{y^2}{4} - \frac{y}{2} - 2 \right) dy$

$$= \left[ \frac{y^3}{12} - \frac{y^2}{4} - 2y \right]_{-2}^{4}$$ $$= \left( \frac{64}{12} - \frac{16}{4} - 8 \right) - \left( \frac{-8}{12} - \frac{4}{4} + 4 \right) = \left( \frac{16}{3} - 4 - 8 \right) - \left( \frac{-2}{3} - 1 + 4 \right) = \left( \frac{16 - 36}{3} \right) - \left( \frac{ -2 + 9 }{3} \right) = \left( \frac{-20}{3} \right) - \left( \frac{7}{3} \right) = \frac{-27}{3} = -9$$

Area is positive, so final answer: $\boxed{9}$