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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