How does a hyperbola go from $(x-h)+(y-k)=-n$ to $(x-h)/a^2 + (y-k)/b^2 = ±1$?

Question

The question is pretty much in the title but to be more specific, actual question is:

$x^2+12x-3y^2+18y+21 = 0$

$(x+6)^2-36-3(y-3)^2+12 = 0$

$(x+6)^2-3(y-3)^2 = -12$

This equation looks a lot like an ellipse. How do I tell if it's an ellipse or a hyperbola? How do I turn this into the form in the title?

What's the issue in dividing both sides by $-12$? – Amadeus Oct 03 '20 at 13:51 — Amadeus, Oct 03 '20 at 13:51
no issue, but how would I get +/- 1 at the end? – John Liu Oct 03 '20 at 14:00 — John Liu, Oct 03 '20 at 14:00

score 1 · Accepted Answer · answered Oct 03 '20 at 14:07

$(x+6)^2-3(y-3)^2 = -12$

$\Rightarrow \frac{(x+6)^2-3(y-3)^2}{12}=-1$

$\Rightarrow \frac{(x+6)^2}{12}-\frac{3(y-3)^2}{12}=-1$

$\Rightarrow \frac{(x+6)^2}{(√12)^2}-\frac{(y-3)^2}{2^2}=-1$

$\therefore h=-6, k=3, a=√12, b=2$

EDIT: I only noticed now that the equation of hyperbola you have mentioned is wrong, $(x-h)/a^2 + (y-k)/b^2 = 1$ is an ellipse, $(x-h)/a^2 - (y-k)/b^2 = ±1$ is a hyperbola

How does a hyperbola go from $(x-h)+(y-k)=-n$ to $(x-h)/a^2 + (y-k)/b^2 = ±1$?

1 Answers1