The intersection of the line and curve can be found by substituting y=-(4+2x) into the first equation:
16+16x+4x^2+16+8x-8x-20=0; 4x^2+16x+12=0; x^2+4x+3=0; (x+1)(x+3)=0, so x=-1 and -3. The y values for these values are y=-(4-2)=-2 and -(4-6)=2. So the intersection points are; (-1,-2) and (-3,2).
Differentiating: 2ydy/dx-4dy/dx-8=0. At (-1,-2) the gradient is given by dy/dx in: -4dy/dx-4dy/dx-8=0; -8dy/dx=8, so dy/dx=-1; at (-3,2), however, 4dy/dx-4dy/dx-8=0, and dy/dx cancels out, leaving the inequality -8=0, so there is apparently only one useful intersection, and the gradient is -1. The tangent line is of the form y=mx+c, where m=dy/dx, the slope, is -1. To find c, insert the point (-1,-2): -2=1+c, making c=-3 and y=-x-3 is the tangent line.
The tangent at (-3,2) is infinity, which explains why we can't get a gradient. The tangent line is, in fact, the line x=-3, and the general equation for this line cannot be covered by the standard equation y=mx+c, because x=-3 for all values of y. So there are two tangent lines: y=-3-x at (-1,-2) and x=-3 at (-3,2). The curve is a sideways parabola with vertex at (-3,2), where the gradient is infinite (the angle is 90 degrees).