[SOLVED] Conic section what each equation represents

**xterminal01** · November 12th 2007, 12:39 AM

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

**earboth** · November 12th 2007, 01:10 AM

Originally Posted by xterminal01

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

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

That's a parabola opening down.

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

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

The ##3 to 5 should be done similary.(H, H, E)

**xterminal01** · November 12th 2007, 01:15 AM

Thanks alot for the help can you please state the step by step for number 3,4,5

**earboth** · November 12th 2007, 01:30 AM

Originally Posted by xterminal01

Thanks alot for the help can you please state the step by step for number 3,4,5

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:

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

to #4:

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

to #5:

$x^2+4y^2=8$ Divide by 8. Then you get the equation of an ellipse.

**xterminal01** · November 13th 2007, 07:03 PM

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

4 i dont know how to finish

5 should be $x^2/8+y^2/2=1$

**Soroban** · November 13th 2007, 08:40 PM

Hello, xterminal01!

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

$1)\;\;x^2-6x+y\:=\:8$
$2)\;\;3x^2+5y^2+6x-10y\:=\:16$
$3)\;\;2x^2+8x\:=\:2y^2-y+10\quad\Rightarrow\quad 2x^2+8x-2y^2+y\:=\:10$
$4)\;\;3x^2+x-y^2+y\:=\:12$
$5)\;\;x^2+4y^2\:=\:8$

If we are to identity the type of conic only , there is an "eyeball" method.

First, get all the variables on one side of the equation.
. . (As we did in #3.)

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

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

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

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

**earboth** · November 13th 2007, 08:48 PM

Originally Posted by xterminal01

3 i got stuck with this...
$2(x+2)^2 -2(x+1/4)^2 = 143/8$

4 i dont know how to finish

5 should be $x^2/8+y^2/2=1$

Hello,

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

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

#4: I'll continue:

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

#3: Divide by $\frac{143}{8}$ to get a 1 at the RHS.

**xterminal01** · November 13th 2007, 08:54 PM

Yea but i cant solve the equation to the end ...

**topsquark** · November 14th 2007, 04:45 AM

Originally Posted by xterminal01

Yea but i cant solve the equation to the end ...

You have all the steps and all but the last one, the division, is given to you. What more do you need?

-Dan

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