# Find Centre of symmetry of conic

• Feb 1st 2013, 06:24 PM
mcleja
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
• Feb 1st 2013, 07:57 PM
MINOANMAN
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
• Feb 1st 2013, 08:18 PM
chiro
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.
• Feb 2nd 2013, 10:37 AM
MINOANMAN
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
• Feb 2nd 2013, 05:25 PM
Soroban
Re: Find Centre of symmetry of conic
Hello, mcleja!

Quote:

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

This is conic is a parabola ... it does not have a center.
It has a vertex.

There are two basic forms for the parabola.

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

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

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

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

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

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