__a=1/2
__b=-8
__a=-1/2
__b=8
__a=-20
Well, given the parabola $\displaystyle y = ax^2$, what is the equation of the line tangent to it at x = 4?
The tangent line must pass through the point (4, 16a).
$\displaystyle y^{\prime} = 2ax$, so when x = 4 the slope of the tangent line is $\displaystyle 2 \cdot a \cdot 4 = 8a$.
So the equation of the line that passes through the point (4, 16a) that has a slope of 8a is:
$\displaystyle y = mx + b$
$\displaystyle y = (8a)x + b$
Insert the point (4, 16a) to find b:
$\displaystyle 16a = (8a)(4) + b$
$\displaystyle b = 16a - 32a = -16a$
So the tangent line is $\displaystyle y = (8a)x - (16a)$ or in the form given in the problem:
$\displaystyle -(8a)x + y = -16a$
Compare this with
$\displaystyle 4x + y = b$
What are your values for a and b?
-Dan