Answer
\[y = \frac{1}{3}x + \frac{5}{3}\]
Work Step by Step
\[\begin{gathered}
\,{\left( {{x^2} + {y^2}} \right)^2} = \frac{{25}}{4}x{y^2}\,\,\,\,\,,\,\,\,\left( {1,2} \right) \hfill \\
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implicit\,\,differentiation \hfill \\
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2\,\left( {{x^2} + {y^2}} \right)\,\left( {2x + 2y{y^,}} \right) = \frac{{25}}{2}xy{y^,} + \frac{{25}}{4}{y^2} \hfill \\
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multiply \hfill \\
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4x\,\left( {{x^2} + {y^2}} \right) + 4y{y^,}\left( {{x^2} + {y^2}} \right) = \frac{{25}}{2}xy{y^,} + \frac{{25}}{4}{y^2} \hfill \\
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collect\,\,like\,\,terms \hfill \\
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4y{y^,}\left( {{x^2} + {y^2}} \right) - \frac{{25}}{2}xy{y^,} = \frac{{25}}{4}{y^2} - 4x\,\left( {{x^2} + {y^2}} \right) \hfill \\
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solve\,\,for\,\,{y^,} \hfill \\
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{y^,} = \frac{{\frac{{25}}{4}{y^2} - 4x\,\left( {{x^2} + {y^2}} \right)}}{{4y\,\left( {{x^2} + {y^2}} \right) + \frac{{25}}{2}xy}} \hfill \\
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evaluate\,\,\,\left( {1,2} \right) \hfill \\
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then \hfill \\
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{y^,} = \frac{{\frac{{25}}{4}\,{{\left( 2 \right)}^2} - 4\,\left( 1 \right)\,\left( {{1^2} + {2^2}} \right)}}{{4\,\left( 2 \right)\,\left( {{1^2} + {2^2}} \right) - \frac{{25}}{2}\,\left( 1 \right)\,\left( 2 \right)}} = \frac{1}{3} \hfill \\
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use\,\,the\,\,point\, - \,slope\,\,form \hfill \\
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y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\
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y - 2 = \frac{1}{3}\,\left( {x - 1} \right) \hfill \\
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y - 2 = \frac{1}{3}x - \frac{1}{3} \hfill \\
\hfill \\
y = \frac{1}{3}x + \frac{5}{3} \hfill \\
\hfill \\
\end{gathered} \]
