How to prove that 2(e-c)=bd

Question

The parabola f(x)=x^2+bx+c has vertex "P" and the parabola g(x)=-x^2+dx+e has vertex "Q."

"P" and "Q" are distinct points. The two parabolas also intersect at "P" and "Q." Prove that

2(e-c)=bd

Answer 1

EDIT: Algebra based answer

Any parabola can be written in vertex form, where a(x) = (x-h)² + k, where the vertex is at x=h. At point P, we know the vertex is at x=y such that f(x) = x² + bx + c = (x-y)² + k.

• Expanding this, we see that: x² + bx + c = x² -2xy +y² + k.
• Removing the x² terms, we get: bx + c = -2xy +y² + k
• Matching the powers of x show us that: bx = -2xy
• And thus y = -b/2.

This is a specific case of the so-called vertex formula

 

Now, we know that at P f(y) = g(y), so

• y² +by + c = -y² + dy + e
• Now substitute y = -b/2
• b²/4 - b²/2+ c = -b²/4 - bd/2 + e
• Now consolidate (all the b² cancel out)
• c =  bd/2 + e
• And rearrange into the desired form
• 2(e-c) = bd

 

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I apologize in advance if this course doesn't use calculus. I make explicit use of derivatives in my answer.

 

First, let's look at point P (at x=y).  

 

At P, we know that f'(y) = 0, so 2y + b = 0 and y = -b / 2.

 

Now, we know that at P f(y) = g(y), so

• y² +by + c = -y² + dy + e
• Now substitute y = -b/2
• b²/4 - b²/2+ c = -b²/4 - bd/2 + e
• Now consolidate (all the b² cancel out)
• c =  bd/2 + e
• And rearrange into the desired form
• 2(e-c) = bd