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

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

$${find}\:{the}\:{equation}\:{of}\:{the}\:{tangent}\:{to}\: \\ $$$${the}\:{curve}\:\sqrt{{x}}+\sqrt{{y}}=\sqrt{{a}}\:{at}\:{any}\:{point} \\ $$$$\left({x},{y}\right){on}\:{it}. \\ $$

Answered by mrW1 last updated on 13/Dec/17

(dx/(2(√x)))+(dy/(2(√y)))=0 ⇒(dy/dx)=−(√(y/x)) eqn. of tangent at (x_1 ,y_1 ): ((y−y_1 )/(x−x_1 ))=−(√(y_1 /x_1 )) ⇒y−y_1 +(x−x_1 )(√(y_1 /x_1 ))=0 with (√x_1 )+(√y_1 )=(√a) or y_1 =x_1 +a−2(√(ax_1 ))

$$\frac{{dx}}{\mathrm{2}\sqrt{{x}}}+\frac{{dy}}{\mathrm{2}\sqrt{{y}}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=−\sqrt{\frac{{y}}{{x}}} \\ $$$$ \\ $$$${eqn}.\:{of}\:{tangent}\:{at}\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right): \\ $$$$\frac{{y}−{y}_{\mathrm{1}} }{{x}−{x}_{\mathrm{1}} }=−\sqrt{\frac{{y}_{\mathrm{1}} }{{x}_{\mathrm{1}} }} \\ $$$$\Rightarrow{y}−{y}_{\mathrm{1}} +\left({x}−{x}_{\mathrm{1}} \right)\sqrt{\frac{{y}_{\mathrm{1}} }{{x}_{\mathrm{1}} }}=\mathrm{0} \\ $$$${with}\:\sqrt{{x}_{\mathrm{1}} }+\sqrt{{y}_{\mathrm{1}} }=\sqrt{{a}} \\ $$$${or} \\ $$$${y}_{\mathrm{1}} ={x}_{\mathrm{1}} +{a}−\mathrm{2}\sqrt{{ax}_{\mathrm{1}} } \\ $$

Commented by mrW1 last updated on 13/Dec/17

you are right sir. thanks!

$${you}\:{are}\:{right}\:{sir}.\:{thanks}! \\ $$

Commented by rita1608 last updated on 13/Dec/17

(dy/dx) is −((√y)/(√x)) plss check ans once again

$$\frac{{dy}}{{dx}}\:{is}\:−\frac{\sqrt{{y}}}{\sqrt{{x}}}\:\:{plss}\:{check}\:{ans}\:{once}\:{again} \\ $$

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