Question Number 170997 by Mastermind last updated on 06/Jun/22
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{tangent}\:{to}\:{the} \\ $$$${curve}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6}{y}−\mathrm{12}=\mathrm{0}\:{at}\:{the} \\ $$$${point}\:\left(\mathrm{2},\:\mathrm{3}\right). \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by depressiveshrek last updated on 06/Jun/22
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6}{y}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{2}{x}+\mathrm{2}{y}\frac{{dy}}{{dx}}−\mathrm{4}+\mathrm{6}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\left(\mathrm{2}{y}+\mathrm{6}\right)\frac{{dy}}{{dx}}+\mathrm{2}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}−\mathrm{2}{x}}{\mathrm{2}{y}+\mathrm{6}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}−\mathrm{2}\centerdot\mathrm{2}}{\mathrm{2}\centerdot\mathrm{3}+\mathrm{6}}=\mathrm{0} \\ $$$${y}={ax}+{b} \\ $$$$\mathrm{3}={b} \\ $$$${y}=\mathrm{3} \\ $$$${The}\:{function}\:{is}\:{undefined}\:{at}\:{y}=\mathrm{3} \\ $$