# Finding the Point of Tangency Theoretically

1. Mar 30, 2006

### SwAnK

Hey I was given the question:
A (x1,y1) and B (x2,y2) are two points on the parabola y=ax^2+bx+c. At what point is the tangent to the parabola parallel to the secant AB.
Here are the steps I took, im just wanting to know if im heading in the right direction.

First I just made the slope of AB = m.
then found the derivative of ax^2+bx+c (2ax+1) and made m=2ax+1.
Then isolated x, which would give an x coordinate of m/2a+1.
Took the x value and subed it into the equation ax^2+bx+c to get a y value.

Is this how you would go about this question?? thanx

2. Mar 30, 2006

### PPonte

Why 2ax + 1, instead of 2ax + b ?

3. Mar 30, 2006

### SwAnK

ahh, yes good point, i wasn't thinking. Thanx for catching that PPonte!

4. Apr 1, 2006

### arildno

Nowhere have you related your "x" to the x1 and x2; what you have done, is to parametrize the parabola in terms of the local slope m which, by the way, you haven't done correctly.

To continue along your track:
In general, you'll get for the tangent slope m
$$m=2ax+b\to{x}=\frac{m-b}{2a}$$

If you now can find the m that is the slope of the secant between x1 and x2, you are done.