UPSC NDA II 2023 Mathematics Solved Paper

Concept:The latus rectum of a parabola $x^{2}=4ay$ has length $4a$.The constant $a$ is found by solving the given differential equation and using the point $(1,2)$.Explanation:Start with the differential equation: $\frac{dy}{dx} = \frac{2y}{x}$, for $x>0, y>0$.Separate variables: $\frac{dy}{2y} = \frac{dx}{x}$.Integrate both sides: $\int \frac{dy}{y} = \int \frac{2\,dx}{x}$.This gives $\ln y = 2\ln x + \ln c$, or $\ln y = \ln(x^{2}c)$.Thus $y = c x^{2}$.The parabola passes through $(1,2)$: substitute $x=1, y=2$ to get $2 = c \cdot 1^{2}$, so $c=2$.Hence the equation is $y = 2x^{2}$, which can be written as $x^{2} = \frac{y}{2}$.Compare with the standard form $x^{2} = 4ay$: here $4a = \frac{1}{2}$.Therefore the length of the latus rectum is $4a = \frac{1}{2}$.Answer:$\frac{1}{2}$ (Option B).