Equation of the normal line?

Question

Find the y-intercept of the normal to the parabola y=x^2 at the point (a,a^2), where a =/= 0.

I need to check an answer. Here's: what I did:

y'=2x, so slope of normal line is -1/2x
Plug in a so the slope is now -1/2a
Equation of the normal line is now y = -1/2ax + b
Plug in the point (a,a^2) so equation is now a^2 = -1/2a(a) + b
Solve for b, so b = 3/2a^2

Did I do that right?

BTW would the limit of that intercept as a -> 0 be equal to 0? You just plug the 0 for a right?

I see what I did wrong now. ty

Answer 1

The slope of the tangent is 2x, so the slope of the normal line is -1/(2x), not -1/2 x, which you used later.
If you continue of, you get -1/(2a), so the equation is y = -1/(2a) * x + b.
So a^2 = -1/(2a) * a + b = -1/2 + b.
Thus b = a^2 + 1/2.

In this case, theres no dividing by a, so you can just substitute in a=0 if you wanted (that was a good hint that your answer was wrong).

Answer 2

looks good to me