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Question: The min. distance between y – x = 1 and y<sup>2</sup> = x is –...

The min. distance between y – x = 1 and y2 = x is –

A

342\frac{3}{4\sqrt{2}}

B

322\frac{3}{2\sqrt{2}}

C

142\frac{1}{4\sqrt{2}}

D

122\frac{1}{2\sqrt{2}}

Answer

342\frac{3}{4\sqrt{2}}

Explanation

Solution

2ydydx\frac{dy}{dx}= 1 Ž dydx\frac{dy}{dx}= 12y\frac{1}{2y} = 1 Ž y = 12\frac{1}{2} & x = 14\frac{1}{4}

Now ⊥ distance from (1/4, 1/2) on

– x + y – 1 = 0 is d = ∣–1/4+1/2–12∣=3/42\left| \frac{–1/4 + 1/2–1}{\sqrt{2}} \right| = 3/4\sqrt{2}