Let , so the second equation becomes
The discriminant of this quadratic is
For the quadratic to even have real solutions for m, the discriminant must be non-negative, i.e. . No calculus needed.
For the other question, differentiate both sides of the original equation:
If the line is a tangent at the point (x,y), it means that x = -6 and dy/dx is undefined, i.e. the denominator of dy/dx is zero. This occurs only when . So try substituting (1,6) and (-1,6) into your original equation and find k.