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I know, from calculating $\delta$ and $\Delta$, that the following function represents parabola.

$$x^2-2xy+y^2-2x+27y+10=0$$

But how can I get it to its canonical form: $y^2=2px$ ?

Essentially I need to sketch it, not precisely draw it.

tyr
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  • Do you just need a sketch of the parabola, or do you need all the math that led to it? Do you know about diagonalization of matrices, and/or eigenvectors? Do you know about "completing the square"? Answers to these questions will help us to help you. – John Hughes Mar 26 '17 at 11:32
  • I do need the math behind it. I know about completing the square. – tyr Mar 26 '17 at 11:33
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    http://www.sosmath.com/CBB/viewtopic.php?t=17029 – lab bhattacharjee Mar 26 '17 at 11:35
  • You could use matrices and then work out the affine transformation you need to get to the canonical form.

    As of now, you have the equation $p(x,y)=x^2-2xy+y^2+27y+10$; your parabola is essentially $P={(x,y)\in \mathbb{R^2} | p(x,y)=0}$. If we let $A=\begin{pmatrix} 1 & -1 & 0 \ -1 & 1 & \frac{27}{2} \ 0 & \frac{27}{2} & 10\end{pmatrix}$, then $P={(x,y,1)\in\mathbb{R^2} | (x,y,1)A\begin{pmatrix} x \ y \ 1\end{pmatrix} = 0}$.

    You want $M$ such that $M^tAM=B$ and $y^2-2px=(x,y,1)B\begin{pmatrix} x \ y \ 1\end{pmatrix} = 0$. M represents the affine transformation you need.

     – AlexT Mar 26 '17 at 11:49

1 Answers1

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OK. Well, your first three terms are just $(x-y)^2$. That suggests changing variables to $$ u = (x+y) \\ v = (x - y) $$ which means that $$ x = (u+v)/2\\ y = (u - v)/2, $$ so that you have $$ v^2 - 2x + 2y + 25y + 10 = 0\\ v^2+2 (y-x) + 25 y + 10 = 0\\ v^2 - 2v + 25 (u-v)/2 + 10 = 0 \\ $$ That's now something for which you can do the completing-the-square to get an aligned parabola.

John Hughes
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