Question

Determine a equação da reta tangente à curva x2+x2y+y2=3 no ponto 1,2

Answer 1

$\large\boxed{\begin{array}{l}\sf x^2+x^2y+y^2=3\\\sf 2x+2xy+x^2\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0\\\sf x^2\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=-2x-2xy\\\sf \dfrac{dy}{dx}(x^2+2y)=-2x-2xy\\\\\sf\dfrac{dy}{dx}=\dfrac{-2x-2xy}{x^2+y}\\\\\sf\dfrac{dy}{dx}\bigg|_{x=1~y=2}=\dfrac{-2\cdot1-2\cdot1\cdot2}{1^2+2}\\\\\sf\dfrac{dy}{dx}\bigg|_{x=1~y=2}=-2\implies f'(1,2)=-2\\\sf y=y_0+f'(1,2)\cdot(x-x_0)\\\sf y=2-2\cdot(x-1)\\\sf y=2-2x+2\\\sf y=-2x+4\longrightarrow \bf equac_{\!\!,}\tilde ao\,da\,reta\,tangente\end{array}}$