Question

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)

Answer 1

`\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(dy)/(dx) \right)-2y(dy)/(dx)+1=0 \\\\\\ 2x+2y+2x(dy)/(dx)-2y(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}\\\\ -------------------------------\\\\`


`\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}}`