find the equation of the tangent to the curve (√x)+(√y)=(√a) at any point (x,y)on it.


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Question Number 25634 by rita1608 last updated on 12/Dec/17

$f i n d t h e e q u a t i o n o f t h e t a n g e n t t o$ $t h e c u r v e \sqrt{x} + \sqrt{y} = \sqrt{a} a t a n y p o i n t$ $(x, y) o n i t .$
	Answered by mrW1 last updated on 13/Dec/17

	$\frac{d x}{2 \sqrt{x}} + \frac{d y}{2 \sqrt{y}} = 0$ $\Rightarrow \frac{d y}{d x} = - \sqrt{\frac{y}{x}}$ $e q n . o f t a n g e n t a t (x_{1}, y_{1}) :$ $\frac{y - y_{1}}{x - x_{1}} = - \sqrt{\frac{y_{1}}{x_{1}}}$ $\Rightarrow y - y_{1} + (x - x_{1}) \sqrt{\frac{y_{1}}{x_{1}}} = 0$ $w i t h \sqrt{x_{1}} + \sqrt{y_{1}} = \sqrt{a}$ $o r$ $y_{1} = x_{1} + a - 2 \sqrt{{a x}_{1}}$
	Commented by mrW1 last updated on 13/Dec/17

	$y o u a r e r i g h t s i r . t h a n k s!$
	Commented by rita1608 last updated on 13/Dec/17

	$\frac{d y}{d x} i s - \frac{\sqrt{y}}{\sqrt{x}} p l s s c h e c k a n s o n c e a g a i n$
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