UPSC NDA II 2024 Mathematics Paper

Concept:The area bounded by a parabola and its latus rectum is computed using definite integration after determining the parabola’s equation from given data.Explanation:The area bounded by $y^2 = kx$ and the line $x = k$ (with $k>0$) is given as $\frac{4}{3}$ square units.This area is symmetric about the x‑axis, so it equals $2\int\limits_0^k y\,dx = 2\int\limits_0^k \sqrt{kx}\,dx$.Evaluate the integral: $2\sqrt{k} \int\limits_0^k x^{\frac{1}{2}}\,dx = 2\sqrt{k}\cdot\frac{2}{3}k^{3/2} = \frac{4}{3}k^2$.Set this equal to $\frac{4}{3}$: $\frac{4}{3}k^2 = \frac{4}{3} \Rightarrow k^2 = 1$.Since $k>0$, $k=1$. Hence the parabola is $y^2 = x$.For $y^2 = x$, we have $4a = 1$, so $a = \frac{1}{4}$.The latus rectum of this parabola is the vertical line $x = a = \frac{1}{4}$.The area bounded by the parabola and its latus rectum is $2\int\limits_0^{\frac{1}{4}} y\,dx = 2\int\limits_0^{\frac{1}{4}} \sqrt{x}\,dx$.Integrate: $2\cdot\frac{2}{3}x^{\frac{3}{2}}\Big|_0^{\frac{1}{4}} = \frac{4}{3}\left(\frac{1}{4}\right)^{\frac{3}{2}} = \frac{4}{3}\cdot\frac{1}{8} = \frac{1}{6}$ square units.Answer:Option A: $\frac{1}{6}$ square unit.