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I have equation $y = -x^2 + 2x + 7$. How can I change it to canonical form, which looks like $y^2 = 2px$ ? ($p$ will be parameter)

What i ve tried so far: $$\begin{align} y &= -x^2 + 2x + 7\\ y &= -(x^2 - 2x + 1) + 8\\ (y-8) &= -(x-1)^2 \\ (y-8)^2 &= 2*(0.5)*(x-1)^4 \end{align} $$

But I have read somewhere its wrong, so how do I make it correct?

Or is my solution correct?

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I have edited and reformatted your question. Please read the faq or see here: meta.math.stackexchange.com/questions/5020/… for questions about how to use LaTeX (MathJax). –  Thomas Jan 19 '13 at 16:27

2 Answers 2

up vote 0 down vote accepted

$$-y=x^2-2x-7=(x-1)^2-8\implies (x-1)^2=-(y-8)$$ which is of the form $(x-\alpha)^2=-4a(y-\beta)$

Now, if we are allowed to make the transformation of axes, we can set $x-1=Y,y-8=X$

So,$Y^2=-X=2\left(-\frac12\right)X$

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well i need it strictly y^2=2px , but from your comment i can say i ve probably done my version right –  user1849353 Jan 19 '13 at 16:27
    
@user1849353, the Right Hand Side in your method is of $O(x^4),$ right? But we need $O(x)$ –  lab bhattacharjee Jan 19 '13 at 16:29
    
Well (x−1)^2=−(y−8) only looks like y^2=2px after you flip x and y by defining X and Y ... is it really the only way ? –  user1849353 Jan 19 '13 at 17:04
    
@user1849353, I think so. –  lab bhattacharjee Jan 19 '13 at 17:31
    
allrightey, many thanks –  user1849353 Jan 19 '13 at 17:58

$$y = -x^2 + 2x + 7 $$ $$y = -(x^2 - 2x +1)+8 $$ $$y = -(x- 1)^2+8 $$ $$(x- 1)^2=-(y-8) $$ $$(x- 1)^2=2(-\frac{1}{2})(y-8)\Rightarrow p=-\frac{1}{2},x-1=Y,y-8=X $$ $$Y^2=2pX$$

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