Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I have edited and reformatted your question. Please read the faq or see here:… for questions about how to use LaTeX (MathJax). – Thomas Jan 19 '13 at 16:27
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 $x-1=Y,y-8=X$


share|cite|improve this answer
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.