Tangent slope

We are given the equation $\large 2x^2+xy+2y^2=4$ . The equation is implicitly defined, thus we will have to implicitly differentiate it.

We start with $\large 2x^2+xy+2y^2=4$

Then, after differentiating the equation we get $\large 4x+y+x\frac { dy }{ dx } +4y\frac { dy }{ dx } =0$

Rearranging the terms leads us to $\large \frac { dy }{ dx } =\frac { -4x-y }{ x+4y }$

We were also given the point $\large (16,-8)$ . To find the slope of the equation at this point we simply plug in the values into the derivative.

That gives us $\large { \frac { dy }{ dx } }_{ (16,-8) }=\frac { -4(16)-(-8) }{ (16)+4(-8) } =\frac { 7 }{ 2 } =3.5$