# Thread: [SOLVED] Conic section what each equation represents

1. ## [SOLVED] Conic section what each equation represents

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

2. Originally Posted by xterminal01
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:
$
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)

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

4. 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.

5. 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 $x^2/8+y^2/2=1$

6. 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)

7. 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.

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

9. 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