Finding the Perpendicular Tangent on a Parabola

[emma3001]

Find the slope of the tangent to the parabola y=-3x^2 + 4x - 7 when x=a. I know how to get the limit using the tangent so i end up with a slope in terms of a (you can also get it using the derivative) but now the next part states:
At what point on the parabola is the tangent perpendicular to the line 3x - 4y + 8=0 and all that i get from that question is that the tangent for the point on the parabola will have a slope of -4/3 (negative reciprocal)

 

[Pere Callahan]

so you have the derivative in terms of a, right? Let's call it D(a).

You're looking for the value for a such that this expression is equal to -4/3.

So what about solving D(a)=-4/3 for a ..?

 

[emma3001]

sorry... so if the derivative is 6x + 4, then i can say that 6x + 4= -4/3 and solve for a (or x)?

 

[Pere Callahan]

Exactly, sometimes it's easier than you think