0
$\begingroup$

In my textbook I am told about a conic section with equation:

$$9x^2 -4y^2 -54x - 16y + 29 = 0$$

and it mentions in passing that since the $xy$ term is $0$ then the centre is:

$\frac{54}{2\times9}, \frac{16}{-2\times 4}$

Presumably they are implicitly using a rule that looks like:

$α = \frac{-D}{2A}, β=\frac{-E}{2C}$ where $(α,β)$ is the centre of a conic section with general form

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

But I've not come across a rule like that. So my question is firstly, am I even inferring the correct rule? And secondly, why does this work?

$\endgroup$
7
  • 1
    $\begingroup$ It's just completing the squares. You should get: $9(x-3)^2-4(y+2)^2=G,$ for some constant $G.$ $\endgroup$ Aug 12, 2021 at 16:15
  • $\begingroup$ And your formula is correct. when $B=0.$ $\endgroup$ Aug 12, 2021 at 16:17
  • $\begingroup$ (Technically, the conic is when that expression $=0.$ The expression itself is not a conic. It is just a quadratic polynomial in two variables.) $\endgroup$ Aug 12, 2021 at 16:19
  • $\begingroup$ @ThomasAndrews I see that now, but then I don't understand why a conic with equation A(x-α)^2 +C(y-β)^2 + F - α^2 - β^2 = 0 has centre (α, β) $\endgroup$
    – physBa
    Aug 12, 2021 at 16:24
  • $\begingroup$ Because if you reflect points around $(\alpha,\beta)$ then you get other points. $(\alpha+a,\beta+b)$ is a point in your conic if and only if any other $(\alpha\pm a,\beta\pm b)$ is on your conic. It definitely depends on your definition of "center of a conic," but that property is key. The center of $Au^2+Cv^2=G$ is $(0,0)$ and this is just a translation of that equation. $\endgroup$ Aug 12, 2021 at 16:27

0

You must log in to answer this question.

Browse other questions tagged .