Question

Pre cal assignment i need help!!!
posting picture in the comments

Answer 1

Circle: \(\color{#000000 }{ \displaystyle (x+h)^2+(y+k)^2=r^2 }\)

Ellipse: \(\color{#000000 }{ \displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 }\)

Hyperbola: \(\color{#000000 }{ \displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 }\)

Parabola: \(\color{#000000 }{ \displaystyle y=ax^2+bx+c }\) or \(\color{#000000 }{ \displaystyle x=ay^2+by+c }\)

Answer 2

those are the "forms" of circle, ellipse, hjyperbola, and parabola.

Answer 3

oh okay

Answer 4

For \(x^2-y^2=9\) divide both sides by 9, (and use the fact that \(9=3^2\).


For \(x=y^2\), you can re-write it as \(x=y^2+0x+0\), and it therefore is a ...?


For \(x^2+y^2=16\), you can re-write it based on the fact that:
\(x^2=(x+0)^2\) and \(y^2=(y+0)^2\) , and you know \(16=4^2\).


For \(4x^2+9y^2=36\), divide both sides by 36, and then you should be able to give me the answer based on the fact that: \(9=3^2\) and \(4=2^2\).

Answer 5

those are the hints/tips.

Answer 6

Okay i just dont get how your supposed to solve these?

Answer 7

you just need to rearrange each equation (as I hinted), and you should end up with one of the above forms for each equation.

Answer 8

hmmmm okay

Answer 9

for x^2+ y^2=16 it would be parabola ?

Answer 10

how did you obtain this result?

Answer 11

I really dont know i just looked at what you said

Answer 12

Possible cases for parabola:
\(\color{#000000 }{ \displaystyle y=ax^2+bx+c }\)
\(\color{#000000 }{ \displaystyle x=ay^2+by+c }\)

(notice that either x is squared and y is not, or the other way around.)

Answer 13

x^2-y^2=9 is parabola

Answer 14

Ok, how did you get this?

Answer 15

from what you posted above i just looked at what was similar

Answer 16

I have an impression that you are trying to guess the answer.

Answer 17

Also, I anyway have to leave right now. I hope you can find someone who will finish it up. Sorry for inconvenicen.