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Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):
A2
B$2\sqrt 2$
C4
D$\sqrt 8$
Answer & Solution
Correct answer: B. $2\sqrt 2$
Minimise $d^2 = x^2 + y^2$ subject to $xy = 4$. By AM-GM: $x^2 + y^2 \ge 2xy = 8$ ⇒ $d \ge \sqrt 8 = 2\sqrt 2$. Equality when $x = y = 2$ (so point is $(2, 2)$, distance $\sqrt 8 = 2\sqrt 2$).
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