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Math Help - [SOLVED] Conic section what each equation represents

  1. #1
    Junior Member xterminal01's Avatar
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    [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
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  2. #2
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    Quote Originally Posted by xterminal01 View Post
    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)
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  3. #3
    Junior Member xterminal01's Avatar
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    Thanks alot for the help can you please state the step by step for number 3,4,5
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  4. #4
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    Quote Originally Posted by xterminal01 View Post
    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.
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  5. #5
    Junior Member xterminal01's Avatar
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    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
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  6. #6
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    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)

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  7. #7
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    Quote Originally Posted by xterminal01 View Post
    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.
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  8. #8
    Junior Member xterminal01's Avatar
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    Yea but i cant solve the equation to the end ...
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by xterminal01 View Post
    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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