Finding a tangent line

**chaddy** · February 6th 2008, 10:24 AM

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)?

**earboth** · February 6th 2008, 11:08 AM

Originally Posted by chaddy

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:

$2x^2 +2xy-y^2=2~\implies~4x+2(y+x \cdot y')-2y \cdot y'=0$ Solve the last equation for y':

$4x+2y+2x \cdot y' - 2y \cdot y'=0~\iff~y'(2x-2y)=-4x-2y~\iff~ y' = \frac{-4x-2y}{2x-2y}$ $= -\frac{2x+y}{x-y}$

Now plug in the coordinates of the tangent point to calculate the slope of the tangent:

$y' = m = -\frac{2+2}{1-2}=4$

Use the point-slope-formula to calculate the equation of the tangent line:

$y-2=4(x-1)~\iff~\boxed{y=4x-2}$

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