the tangent to thecurve 3xy^2-2x^2y=1 at(1,1) meets the curve again at which point?

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ 3{\mathrm{xy}}^2-2{\mathrm{x}}^2\mathrm{y}=1 \\ \mathrm{Differentiate} \\ 3\left[\mathrm{x}.2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{y}}^2\right]-2\left[2\mathrm{x}.\mathrm{y}+{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}\right]=0 \\ \mathrm{Put} \mathrm{x}=\mathrm{y}=1 \\ 3\left[2\frac{\mathrm{dy}}{\mathrm{dx}}+1\right]-2\left[2+\frac{\mathrm{dy}}{\mathrm{dx}}\right]=0 \\ 6\frac{\mathrm{dy}}{\mathrm{dx}}+3-4-2\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ 4\frac{\mathrm{dy}}{\mathrm{dx}}=1 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{4} \\ \mathrm{y}-1=\frac{1}{4}\left(\mathrm{x}-1\right) \\ \Rightarrow 4\mathrm{y}-4=\mathrm{x}-1 \\ \Rightarrow \mathrm{x}-4\mathrm{y}+3=0 \\ \Rightarrow \mathrm{x}=3-4\mathrm{y} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \mathrm{curve} \\ 3\left(3-4\mathrm{y}\right){\mathrm{y}}^2-2{\mathrm{x}}^2\left(3-4\mathrm{y}\right)=1 \\ \mathrm{Now} \mathrm{solve} \mathrm{to} \mathrm{get} \mathrm{required} \mathrm{point}. \end{gathered}$$

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