Finding the Point of Tangency Theoretically

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To find the point of tangency on the parabola y=ax^2+bx+c where the tangent is parallel to the secant line between points A(x1,y1) and B(x2,y2), the slope of the secant line, m, should be calculated first. The derivative of the parabola, 2ax+b, gives the slope of the tangent line, which should equal m. The correct relationship to isolate x is x=(m-b)/(2a). By determining the slope of the secant line between the two points, the problem can be solved effectively. This approach ensures the tangent line is parallel to the secant line as required.
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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, I am just wanting to know if I am 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
 
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then found the derivative of ax^2+bx+c (2ax+1) and made m=2ax+1.

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

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