UPSC NDA I 2019 Mathematics Solved Paper

Concept:Minimise the distance from a fixed point to a point on a parabola using differentiation.Explanation:Step 1: Parameterise a point on the parabola $y^2 = 8x$.Let the y-coordinate be $t$. Then $x = \frac{t^2}{8}$.So the point is $\left(\frac{t^2}{8}, t\right)$.Step 2: Write the square of the distance $D$ from this point to $(4, 2)$.$D^2 = \left(\frac{t^2}{8} - 4\right)^2 + (t - 2)^2$$= \frac{1}{64}(t^2 - 32)^2 + (t - 2)^2$Step 3: To minimise $D$, minimise $D^2$. Differentiate with respect to $t$ and set derivative to $0$.$\frac{d(D^2)}{dt} = \frac{1}{64} \cdot 2(t^2 - 32) \cdot 2t + 2(t - 2)$$= \frac{1}{16} t(t^2 - 32) + 2(t - 2) = 0$Step 4: Simplify the equation.$\frac{t^3}{16} - 2t + 2t - 4 = 0$$\Rightarrow \frac{t^3}{16} = 4$$\Rightarrow t^3 = 64$$\Rightarrow t = 4$Step 5: Substitute $t = 4$ into the distance squared formula.$D^2 = \frac{1}{64}(16 - 32)^2 + (4 - 2)^2$$= \frac{1}{64}(-16)^2 + (2)^2 = \frac{256}{64} + 4 = 4 + 4 = 8$So $D = \sqrt{8} = 2\sqrt{2}$.Answer:$2\sqrt{2}$ (Option B).