shortest distance from (−6,0)to x^2 −y^2 +16=0


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Question Number 201829 by 281981 last updated on 13/Dec/23

$s h o r t e s t d i s t a n c e f r o m (- 6, 0) t o x^{2} - y^{2} + 16 = 0$
	Answered by esmaeil last updated on 13/Dec/23

	$d = \sqrt{{(x + 6)}^{2} + y^{2}} = \sqrt{\overset{2}{x} + 12 x + 36 + \overset{2}{x} + 16} =$ $Missing \left or extra \right$ $i f (x = - 3) \to d = s h o r t e s t = \sqrt{34}$
	Commented by 281981 last updated on 14/Dec/23

	$t n q s i r$
	Answered by mr W last updated on 13/Dec/23

	${(x + 6)}^{2} + y^{2} = d^{2}$ $x^{2} - y^{2} + 16 = 0$ $x^{2} - d^{2} + {(x^{2} + 6)}^{2} + 16 = 0$ $2 x^{2} + 12 x + 52 - d^{2} = 0$ $Δ = 12^{2} - 4 \times 2 \times (52 - d^{2}) = 0$ $d^{2} = 34$ $\Rightarrow d = \sqrt{34}$
	Commented by 281981 last updated on 14/Dec/23

	$t n q s i r$
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