A hyperbola is given by the equation x^2+2xy−y^2+x = 2. Use implicit differentiation
to find an equation of the tangent line to this curve at the point (1, 2)

Respuesta :

[tex]\bf x^2+2xy-y^2+x=2\qquad \begin{array}{llll} (1&,&2)\\ x_1&&y_1 \end{array}\\\\ -------------------------------\\\\ 2x+2\left( 1\cdot y+x\frac{dy}{dx} \right)-2y\frac{dy}{dx}+1=0 \\\\\\ 2x+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}+1=0 \\\\\\ \cfrac{dy}{dx}(2x-2y)=-1-2x-2y\implies \cfrac{dy}{dx}=\cfrac{-1-2x-2y}{2x-2y} \\\\\\ \left. \cfrac{dy}{dx}=\cfrac{2x+2y+1}{2y-2x} \right|_{1,2}\implies \cfrac{2+4+1}{4-2}\implies \cfrac{7}{2}\\\\ -------------------------------\\\\ [/tex]

[tex]\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=\cfrac{7}{2}(x-1)\implies y-2=\cfrac{7}{2}x-\cfrac{7}{2} \\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ \boxed{y=\cfrac{7}{2}x-\cfrac{3}{2}}[/tex]

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