Answer
$(y-1)=-\frac{1}{2}(x-2)$
Work Step by Step
$(x^2+2xy+4y^2=12)'$
Using the chain rule:
$2x+2xy'+2y+8yy'=0$.
Plug in the given point, $(2, 1)$, and then solve for $y'$, which is the slope.
$2(2)+2(2)y'+2(1)+8(1)y'=0$
$6+12y'=0$
$y'=m=-\frac{1}{2}$
Using point-slope form, we get:
$(y-1)=-\frac{1}{2}(x-2)$
