Find Centre of symmetry of conic

I have the following conic and i want to find its centre of symmetry: 3y^2+2x+18y=-31

I tried the following way but it didn't seem to work, could somebody tell me where i went wrong?

(3y^2+18y)+2x=-31

3(y^2+6y)+2x=-31

completing the square:

3(y^2+6y+9)+2x=-31+(3*9)

3(y^2+6y+9)+2x=-4

3(y+3)^2+2x=-4

(3(y+3)^2)/(-4)+2x/(-4)=1

so the centre is (0,-3)?

Where am i going wrong?

Thanks

Re: Find Centre of symmetry of conic

The conic you mentioned above is a shifted parabola .

the full formula is (y+3)^2=4(-1/6)(x+2) and its centre lies at the point P(-2,-3).

a shifted parabola of this type has the formula (y+b)^2=4p(x+a) with p the parameter of the parabola and centre at P(-a,-b)

Minoas

M.R

Re: Find Centre of symmetry of conic

Hey mcleja.

Your working looks good but I don't think this is a normal conic since you will get many complex solutions.

This also doesn't have the form of a circle, ellipse, or a hyperbola.

Re: Find Centre of symmetry of conic

As I said before this is a shifted Parabola. you may refere to any calculus book for further reference about shifted conics.

As a general rule if a conic has the quadratic only in one variable like x or y only then it is a parabola.

Minoas

Re: Find Centre of symmetry of conic

Hello, mcleja!

Quote:

I have the following conic and i want to find its centre of symmetry:

. . $\displaystyle 3y^2+2x+18y\:=\:-31$

This is conic is a *parabola* ... it does not have a center.

It has a vertex.

There are two basic forms for the parabola.

$\displaystyle \begin{Bmatrix}\text{Vertical }(\cup\text{ or }\cap): & (x-h)^2 \:=\:4p(y-k) \\ \text{Horizontal }(\subset\text{or}\supset): & (y-k)^2 \:=\:4p(x-h) \end{Bmatrix}\;\;\text{where }(h,k)\text{ is the vertex.}$

We have: .$\displaystyle 3y^2 + 2x + 18y \:=\:\text{-}31 \quad\Rightarrow\quad 3y^2 + 18y \:=\:\text{-}2x - 31$

. . . . . . . .$\displaystyle 3(y^2 + 6y) \:=\:\text{-}2x-31 \quad\Rightarrow\quad 3(y^2 + 6y + 9) \:=\:\text{-}2x - 31 + 27$

. . . . . . . .$\displaystyle 3(y+3)^2 \:=\:\text{-}2x - 4 \quad\Rightarrow\quad 3(y+3)^2 \:=\:\text{-}2(x+2) $

. . . . . . . .$\displaystyle (y+3)^2 \:=\:\text{-}\tfrac{2}{3}(x+2)$

The parabola opens to the *left* and has vertex $\displaystyle (\text{-}2,\,\text{-}3)$