Can someone please help me with this,

I am suppose to state which type of conic section is represented by each equation step by step if its possible

1. x^2-6x+y=8

2. 3x^2+5y^2+6x-10y=16

3. 2x^2+8x=2y^2-y+10

4. 3x^2+x-y^2+y=12

5. x^2+4y^2=8

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- Nov 12th 2007, 12:39 AMxterminal01[SOLVED] Conic section what each equation represents
Can someone please help me with this,

I am suppose to state which type of conic section is represented by each equation step by step if its possible

1. x^2-6x+y=8

2. 3x^2+5y^2+6x-10y=16

3. 2x^2+8x=2y^2-y+10

4. 3x^2+x-y^2+y=12

5. x^2+4y^2=8 - Nov 12th 2007, 01:10 AMearboth
Hello,

rearrange your equations until you can determine the type of conic. All of them have their axes parallel to the coordinate axes so completing the square will do:

to #1:

$\displaystyle x^2-6x+y = 8~\iff~(x^2-6x+9)+y=8+9~\iff~ y = -(x-3)^2+17$

That's a parabola opening down.

to #2:

$\displaystyle

3x^2+5y^2+6x-10y=16~\iff~3(x^2+2x+1) + 5(y^2-2y+1)=16+3+5~\iff~$ $\displaystyle 3(x+1)^2+5(y-1)^2=24$ . Now divide by 24 and you'll get:

$\displaystyle \frac{(x+1)^2}{8}+\frac{(y-1)^2}{\frac{24}{5}}=1$ That's the equation of an ellipse.

The ##3 to 5 should be done similary.(H, H, E) - Nov 12th 2007, 01:15 AMxterminal01
Thanks alot for the help can you please state the step by step for number 3,4,5

- Nov 12th 2007, 01:30 AMearboth
Hi,

I was quite sure that you could do the last problems using the way I demonstrated to you, but .... I'll start the problems and I'll leave the final brush up to you:

to #3:

$\displaystyle 2x^2+8x=2y^2-y+10~\iff~2x^2+8x-2y^2+y=10~\iff~$ $\displaystyle 2(x^2+4x+4)-2(y^2+\frac12 y+\frac{1}{16})= 10+8-\frac{2}{16}$ . Simplify and you should come out with a hyperbola

to #4:

$\displaystyle 3x^2+x-y^2+y=12~\iff~3(x^2+\frac13 x+\frac{1}{36})-(y^2-y+\frac14)=12 + \frac{3}{36} - \frac14$ Simplify. It's a hyperbola too.

to #5:

$\displaystyle x^2+4y^2=8$ Divide by 8. Then you get the equation of an ellipse. - Nov 13th 2007, 07:03 PMxterminal01
Okey so i tried to finish number 3 and this is what i came up with please check if its right

3 i got stuck with this...

$\displaystyle 2(x+2)^2 -2(x+1/4)^2 = 143/8$

4 i dont know how to finish :confused:

5 should be $\displaystyle x^2/8+y^2/2=1$ - Nov 13th 2007, 08:40 PMSoroban
Hello, xterminal01!

Quote:

State which type of conic section is represented by each equation.

$\displaystyle 1)\;\;x^2-6x+y\:=\:8$

$\displaystyle 2)\;\;3x^2+5y^2+6x-10y\:=\:16$

$\displaystyle 3)\;\;2x^2+8x\:=\:2y^2-y+10\quad\Rightarrow\quad 2x^2+8x-2y^2+y\:=\:10$

$\displaystyle 4)\;\;3x^2+x-y^2+y\:=\:12$

$\displaystyle 5)\;\;x^2+4y^2\:=\:8$

First, get all the variables on one side of the equation.

. . (As we did in #3.)

If it has either $\displaystyle x^2$ or $\displaystyle y^2$,*but not both*, it is a. .(#1)*parabola*

If $\displaystyle x^2$ and $\displaystyle y^2$ have the same coefficient, it is a. .(None listed)*circle*

If $\displaystyle x^2$ and $\displaystyle y^2$ have the same sign, it is an. .(#2 and 5)*ellipse*

If $\displaystyle x^2$ and $\displaystyle y^2$ have opposite signs, it is a. .(#3 and 4)*hyperbola*

- Nov 13th 2007, 08:48 PMearboth
Hello,

you've done #5 correctly. I would have written:

$\displaystyle \frac{x^2}{(2\sqrt{2})^2}+\frac{y^2}{(\sqrt{2})^2} =1$

#4: I'll continue:

$\displaystyle 3(x^2+\frac13 x+\frac{1}{36})-(y^2-y+\frac14)= \frac{71}{6}~\iff~ 3\left(x+\frac16\right)^2-\left(y-\frac12\right)^2=\frac{71}{6}$ and now divide by $\displaystyle \frac{71}{6}$. That's all.

#3: Divide by $\displaystyle \frac{143}{8}$ to get a 1 at the RHS. - Nov 13th 2007, 08:54 PMxterminal01
Yea but i cant solve the equation to the end ...

- Nov 14th 2007, 04:45 AMtopsquark