Answer
a) $m=-2$
b) $y=-2x$
Work Step by Step
We have: $2x^2-y^2=xy$
We differentiate both sides with respect to $x$.
$4x-2y \dfrac{dy}{dx}=x \dfrac{dy}{dx}+y\\ \dfrac{dy}{dx}=\dfrac{4x-y}{x+2y} $
a) The slope at $( -1,2)$ is: $m=\dfrac{4(-1)-2}{-1+4} =-2$
b) The equation of a tangent line at $( -1,2)$ is:
$y-y_1=m(x-x_1)\\ y-2=-2(x+1)\\ y-2=-2x-2\\ y=-2x$
