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Question Number 147905 by puissant last updated on 24/Jul/21

Answered by Olaf_Thorendsen last updated on 24/Jul/21

$$x^2+y^2=13$$$$2xdx+2ydy=0$$$$x+y\frac{dy}{dx}=0$$$$\bullet \mathrm{En}(2,-3):$$$$2-3\frac{dy}{dx}=0$$$$\frac{dy}{dx}=\frac{2}{3}$$$$$$$$\mathrm{Remarque}:x^2+y^2=13\mathrm{est}{\mathrm{l}}^{\prime }\mathrm{equation}$$$$\mathrm{du}\mathrm{cercle}\mathrm{\Omega }\mathrm{de}\mathrm{centre}{\mathrm{l}}^{\prime }\mathrm{origine}\mathrm{O}\mathrm{et}\mathrm{de}$$$$\mathrm{rayon}\mathrm{R}=\sqrt{13}.$$$$\mathrm{A}\mathrm{chaque}\mathrm{point}\mathrm{M}\mathrm{du}\mathrm{cercle}\mathrm{de}$$$$\mathrm{parametre}\theta ,\mathrm{on}\mathrm{peut}\mathrm{associer}\mathrm{le}$$$$\mathrm{rayon}-\mathrm{vecteur}\overrightarrow{\mathrm{OM}}=\left(\begin{array}{c}\mathrm{Rcos}\theta \\ \mathrm{Rsin}\theta \end{array}\right).\mathrm{Sa}$$$$\mathrm{pente}\mathrm{est}\frac{y_{\mathrm{M}}}{x_{\mathrm{M}}}=\tan \theta .$$$$\mathrm{La}\mathrm{tangente}\mathrm{au}\mathrm{cercle}\mathrm{en}\mathrm{M}\mathrm{est}$$$$\mathrm{perpendiculaire}\mathrm{au}\mathrm{rayon}-\mathrm{vecteur}.\mathrm{Sa}$$$$\mathrm{pente}\mathrm{est}\mathrm{donc}-\frac{x_M}{y_{\mathrm{M}}}=-\cot \theta .\mathrm{Ce}\mathrm{qui}$$$$\mathrm{vaut}\mathrm{ici}-\left(\frac{2}{-3}\right)=\frac{2}{3}.$$$$\mathrm{On}\mathrm{retrouve}\mathrm{bien}\mathrm{le}\mathrm{resultat}\mathrm{precedent}.$$

Commented by puissant last updated on 24/Jul/21

$$\mathrm{good}..$$

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