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  • The slope of tangent at any point (x, y) on a curve y = y(x) is   <img alt="\frac{x^{2}+y^{2}}{2xy}\, ,\, x>0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E%7B2%7

The slope of tangent at any point (x, y) on a curve y = y(x) is   \frac{x^{2}+y^{2}}{2xy}\, ,\, x>0  If y (2) = 0, then a value of y (8) is:

Option: 1

4\sqrt{3}


Option: 2

-4\sqrt{2}


Option: 3

-2\sqrt{3}


Option: 4

2\sqrt{3}


Answers (1)

best_answer

\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}\\

y=vx\\y(2)=0\\y(8)=?\\\frac{dy}{sx}=v+x\frac{dv}{dx}

v+\frac{x d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}\\ x \cdot \frac{d v}{d x}=\left(\frac{v^2+1}{2 v}-v\right)\\ \frac{2 v d v}{\left(1-v^2\right)}=\frac{d x}{x}

\begin{aligned} & -\ln \left(1-v^2\right)=\ln x+C \\ & \ln x+\ln \left(1-v^2\right)=C \\ & \ln \left[x\left(1-\frac{y^2}{x^2}\right)\right]=C \end{aligned}

\begin{aligned} & \ln \left[\left(\frac{x^2-y^2}{x}\right)\right]=C \\ & x^2-y^2=c x \\ & y(2)=0 \text { at } x=2, y=0 \\ & 4=2 C \Rightarrow C=2 \\ & x^2-y^2=2 x \end{aligned}

Hence, at x = 8

64-y^{2}=16

\begin{aligned} & y=\sqrt{48}=4 \sqrt{3} \\ & y(8)=4 \sqrt{3} \end{aligned}

Option (1)

 

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