What are the coordinates of the points on the curve x^2 −y^2 =16 which nearest to (0,6)?


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Question Number 157947 by cortano last updated on 30/Oct/21

$W h a t a r e t h e c o o r d i n a t e s o f t h e$ $p o i n t s o n t h e c u r v e x^{2} - y^{2} = 16$ $w h i c h n e a r e s t t o (0, 6) ?$
	Answered by mr W last updated on 30/Oct/21

	$d^{2} = x^{2} + {(y - 6)}^{2}$ $= 16 + y^{2} + {(y - 6)}^{2}$ $= 2 {(y - 3)}^{2} + 34 ⩾ 34$ $d ⩾ \sqrt{34} = n e a r e s t d i s t a n c e$ $a t y = 3, x = \pm \sqrt{16 + 3^{2}} = \pm 5$
	Commented by cortano last updated on 30/Oct/21

	$y e s$
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