Find the shortest distance between point A(3,2) and curve y=(√x) (x>0).


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Question Number 201659 by LimPorly last updated on 10/Dec/23

$F i n d t h e s h o r t e s t d i s t a n c e b e t w e e n$ $p o i n t A (3, 2) a n d c u r v e y = \sqrt{x} (x > 0) .$
	Answered by mr W last updated on 10/Dec/23

	$s a y t h e d i s t a n c e i s s .$ $s a y t h e p o i n t o n t h e c u r v e i s (p^{2}, p)$ $Φ = s^{2} = {(p^{2} - 3)}^{2} + {(p - 2)}^{2}$ $\frac{d Φ}{d p} = 4 p (p^{2} - 3) + 2 (p - 2) = 0$ $p^{3} - \frac{5 p}{2} - 1 = 0$ $\Rightarrow p = \frac{\sqrt{30}}{3} \sin (\frac{π}{3} + \frac{1}{3} \sin^{- 1} \frac{3 \sqrt{30}}{25})$ $s_{m i n} = \sqrt{13 - \frac{75}{9} \sin^{2} (\frac{π}{3} + \frac{1}{3} \sin^{- 1} \frac{3 \sqrt{30}}{25}) - \sqrt{30} \sin (\frac{π}{3} + \frac{1}{3} \sin^{- 1} \frac{3 \sqrt{30}}{25})}$ $\approx 0.257 552 568$
	Commented by mr W last updated on 10/Dec/23

	Commented by LimPorly last updated on 10/Dec/23

	$T h a n k y o u s i r$
	Answered by mr W last updated on 11/Dec/23

	$y = x^{2}$ $A (2, 3)$ $P (p, p^{2})$ $\tan θ = 2 p$ $2 = p + r \sin θ . . . (i)$ $3 = p^{2} - r \cos θ . . . (i i)$ $r^{2} = {(2 - p)}^{2} + {(p^{2} - 3)}^{2}$ $\frac{2 - p}{p^{2} - 3} = 2 p$ $p^{3} - \frac{5 p}{2} - 1 = 0$ $\Rightarrow p = \frac{\sqrt{30}}{3} \sin (\frac{π}{3} + \frac{1}{3} \sin^{- 1} \frac{3 \sqrt{30}}{25})$
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