I need to find the equation of the line tangent to the graph of 2x^2 +2xy-y^2=2 at the point (1,2)?
Use implicite derivation:
$\displaystyle 2x^2 +2xy-y^2=2~\implies~4x+2(y+x \cdot y')-2y \cdot y'=0$ Solve the last equation for y':
$\displaystyle 4x+2y+2x \cdot y' - 2y \cdot y'=0~\iff~y'(2x-2y)=-4x-2y~\iff~ y' = \frac{-4x-2y}{2x-2y}$ $\displaystyle = -\frac{2x+y}{x-y}$
Now plug in the coordinates of the tangent point to calculate the slope of the tangent:
$\displaystyle y' = m = -\frac{2+2}{1-2}=4$
Use the point-slope-formula to calculate the equation of the tangent line:
$\displaystyle y-2=4(x-1)~\iff~\boxed{y=4x-2}$