Question

\[25x ^{2}-{36y^2}+900=0\] ----- I know how to work the equation, but what about the negative polynomial x^2 it leaves, how does that work?

Answer 1

\[25x ^{2}-{36y^2}+900=0\]

Answer 2

what are you trying to do @Jeremy.Kane

Answer 3

Trying to put this in standard hyperbola form?

Answer 4

convert it to a standard ellipse or parabola form, but it leaves me with the first polynomial being a negative and I can't figure out how that changes anything

Answer 5

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]
This is our goal!
It has to be a hyperbola if any of the conics
because the x^2 and y^2 have different signs in front of it

Answer 6

right, the minus sign shows that, so the negative x component doesn't change anything? I have -36^2 at the bottom, then I switch the sign to the top .. I'd get -(x+h)^2 then? that negative doesn't make a difference because it's a square?

Answer 7

sorry, I should use the equation editor

Answer 8

Or goal is really either this \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]
or
this
\[\frac{-(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]


This is our equation
\[25x^2-36y^2=-900 \]

We can tell we want one side to be 1
so that is what I try to do first
the way we get one is by dividing
-900/-900=1

If you divide one side, you must divide the other side by the same thing.

Answer 9

Both of those forms I just mentioned or hyperbola forms

Answer 10

your a mod, I didn't mean to give someone an answer a few minutes ago- I wasn't aware of that rule.

And that is what I was thinking. the \[-(x-0)^{2} \] won't change anything because it's a square

Answer 11

It's fine.

If you are asking me if (-1)^2 is 1 or 1^2, then yes
But -1^2 is not (-1)^2 which means -1^2 is not 1^2

Answer 12

your equation is definitely not an ellipse or a parabola

Answer 13

This is why I was saying if it is any conic it is a hyperbola

Answer 14

http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

Here are some examples of hyperbolas.

Let me know if you need more help if you return.

Answer 15

im sorry, had to check with the kiddo -- it's a conic section for sure, i have to name it and find its vertices and foci - its just that negative first part that is confusing the heck out of me. I come up with

Answer 16

shoot, one sec

Answer 17

\[\frac{ (x-0)^{2} }{ -6^2 }\] which is the same as \[-(x-0)^{2}/{6^2}\] .. its the negative sign that is throwing me off -- really do appreciate this help

Answer 18

I hope that makes sense