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  • Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is \frac{x^{2}+y^{2}}{2xy}

Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is \frac{x^{2}+y^{2}}{2xy}

Answers (1)

Slope of the tangent is given by \frac{x^{2}+y^{2}}{2 x y}$
Slope of the tangent of the curve \mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{d} \mathrm{x}$

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

Put y=VX

\Rightarrow \frac{\mathrm{d}(\mathrm{vx})}{\mathrm{dx}}=\frac{\mathrm{x}^{2}+\mathrm{v}^{2} \mathrm{x}^{2}}{2 \mathrm{vx}^{2}}$

Using product rule differentiate vx

\\\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}=\frac{1+\mathrm{v}^{2}}{2 \mathrm{v}}$ \\$\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}}{2 \mathrm{v}}-\mathrm{v}$ \\$\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}-2 \mathrm{v}^{2}}{2 \mathrm{v}}$ \\$\Rightarrow \frac{d v}{d x}=\left(\frac{1}{x}\right) \frac{1-v^{2}}{2 v}$
$$ \Rightarrow \frac{2 v}{1-v^{2}} d v=\left(\frac{1}{x}\right) d x $$
Integrate
$$ \Rightarrow \int \frac{2 v d v}{1-v^{2}}=\int\left(\frac{1}{x}\right) d x $$
Put 1-\mathrm{v}^{2}=\mathrm{t}$
2 \mathrm{vdv}=-\mathrm{dt}$

$$ \\ \Rightarrow \int \frac{-d t}{t}=\log x+c \\ \Rightarrow-\log t=\log x+c $$
Resubstitute 1
$$ \Rightarrow-\log \left(1-v^{2}\right)=\log x+c $$
Resubstitute v
$$ \Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x+c \ldots(a) $$
The curve is passing through (2,1)
Hence (2,1) will satisfy the equation (a)
Put x=1 and  y=2  in  (a) 

\\\Rightarrow-\log \left(1-\frac{1^{2}}{2^{2}}\right)=\log 2+c$ \\$\Rightarrow-\log \left(\frac{4-1}{4}\right)=\log 2+\mathrm{c}$ \\$\Rightarrow-\log \left(\frac{3}{4}\right)=\log 2+\mathrm{c}$ \\$\Rightarrow-\log \left(\frac{3}{4}\right)-\log 2=c$ \\$\Rightarrow-\left(\log \left(\frac{3}{4}\right)+\log 2\right)=c$
Use loga+logb=logab
\Rightarrow-\log \frac{3}{2}=c$
Put c in equation (a)

\\ \Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x-\log \frac{3}{2} \\ \Rightarrow-\log \left(\frac{x^{2}-y^{2}}{x^{2}}\right)=\log x-\log \frac{3}{2} \\ \Rightarrow \log \left(\frac{x^{2}-y^{2}}{x^{2}}\right)^{-1}=\log x-\log \frac{3}{2} \\ \Rightarrow \log \left(\frac{x^{2}}{x^{2}-y^{2}}\right)-\log x=-\log \frac{3}{2} \\ \Rightarrow \quad \frac{3 x}{2\left(x^{2}-y^{2}\right)}=e^{0} \\ \Rightarrow 3 x=2 x^{2}-2 y^{2}

\begin{aligned} &\Rightarrow 2 y^{2}=2 x^{2}-3 x\\ &\Rightarrow \mathrm{y}=\sqrt{\frac{2 \mathrm{x}^{2}-3 \mathrm{x}}{2}}\\ &\text { Hence the equation of the curve is }&y=\sqrt{\frac{2 x^{2}-3 x}{2}}\\ \end{aligned}

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