given the equation of the curve as y=8-2x^2
(a) find equations of tangent and the normal to this curve at the point where x=a.
(b) find the value of a for which this tangent is parallel to the line with equation 3x-y+6=0
y' = -4x
at x=a, y'=-4a
so, we want the line through (a,8-4a^2) with slope -4a.
y+4a^2-8 = -4a(x-a)
(b) we want -4a to be parallel to the line with slope 3.
So, -4a=3, and a = -3/4