Mathematics: Ahyperbola 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)

Ahyperbola is given by the equation x^2+2xy−y^2+x

Ответ:

danielroytman1

02.01.2022

$\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}\\\\
-------------------------------\\\\
$

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

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Ответ:

pannita

04.02.2022

Mathematics

Step-by-step explanation:

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Ahyperbola 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)

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