Locus of the Vertices of Parabolas

The relationship between y = ax2 + bx + c and y = -ax2 + c

 

By Pei-Chun Shih

 

 

From the previous write-up, we found that the locus of the vertices of the set of parabolas graphed from y = x2 + bx + 1 is the locus of the parabola y = -x2 + 1:

 

[The red parabola, y = - x2 + 1, connects the vertices of the rest parabolas]

 

It looks like there exists relationship between the equation y = ax2 + bx + c and the equation y = -ax2 + c. But before making any conclusion, I will do several more explorations to support my argument.

 


 

y = 2x2 + bx + 1 & y = -2x2 + 1 (let a = 2, c = 1 and vary the value of b):

 

 

The equation y = -2x2 + 1 passes through the following points: (-1, -1), (1, -1), (-0.75, -0.125), (0.75, -0.125), (-0.5, 0.5), (0.5, 0.5), (-0.25, 0.875), (0.25, 0.875), and (0, 1) which are the vertices of the parabolas of equation y = 2x2 + bx + 1, with b = 0, 1, 2, 3, and 4.

 


 

y = -3x2 + bx + 2 & y = 3x2 + 2 (let a = -3, c = 2 and vary the value of b):

 

 

Again, the equation y = 3x2 + 2 passes through the points which are the vertices of the parabolas of equation y = -3x2 + bx + 2, with b = 0, 1, 2, 3, and 4.

 

Therefore, we can generalize the equation of the locus, for any equation of the form y = ax2 + bx + c as the equation of:

y = -ax2 + c.

 


 

Well, generalizing by graphics along is not complete enough for a rigorous proof in mathematics. Therefore, here I am going to use algebra to prove the relationship between the equations of y = ax2 + bx + c and y = -ax2 + c.

 

As I have discussed in my previous write-up 2 that the standard form of the equation of a parabola with a vertical axis of symmetry can be written as (x - h)2 = 4p (y - k), where (h, k) is the vertex, (h, k + p) is the focus, x = h is the axis of symmetry, and y = k – p is the directrix. Since we have known that the equation, y = ax2 + bx + c, represents a parabola with a vertical axis of symmetry, it is reasonable to write it in the standard form of (x - h)2 = 4p (y - k).

 

y = ax2 + bx + c

ax2 + bx = y – c

x2 +x = (y – c)

x2 + 2* x *+ ()2 = (y – c) + ()2

(x+)2 = (y – c+)

(x+)2 = 4*[ y – (c -)]

 

Therefore, h = -, k = c -, and the vertex of y = ax2 + bx + c is (-, c -).

Since we want to prove that every vertex from the equation y = ax2 + bx + c is also on the equation y = -ax2 + c, plug the vertex

(-, c -) into the equation y = -ax2 + c is the next step.

 

y = -ax2 + c

Let x = - and y = c -.

c - = -a*(-)2 + c

c - = -+ c

 

Thus, 0 = 0. This confirms that every point of the form (-, c -) is on the equation y = -ax2 + c.

 

By the algebraic statement above, we have proved that the locus of the vertices of the set of parabolas graphed from

y = ax2 + bx + c is the locus of the parabola y = -ax2 + c.

 

 

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