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Geometry - Conic Sections

  1. Jun 19, 2013 #1
    Hi all,

    I am having an issue with the following problem. I just don't know how to approach it.

    1. The problem statement, all variables and given/known data
    kxL7ZEC.png


    2. Relevant equations
    Ax^2 + Bxy + Cy2 + Dx + Ey + F = 0

    3. The attempt at a solution
    CM3YpiC.png
    WF5z5Ay.png

    I am confused on how to put this problem in terms of x & y and get numerical values for both x & y. Any guidance would be greatly appreciated :)
     
  2. jcsd
  3. Jun 19, 2013 #2

    LCKurtz

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    Try putting ##\sin\alpha = y/r## and ##\cos\alpha = x/r## in your equation.
     
  4. Jun 19, 2013 #3
    Great thank you. That was a big help and is starting to make sense now. Since it is a right triangle, your equations can be used.
    [tex]x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}[/tex]

    [tex]\therefore x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}[/tex]

    [tex]\therefore x^{2}+y^{2} = (\frac{x}{r}) * \frac{y^{2}}{r^{2}}[/tex]

    [tex]\therefore x^{2}+y^{2} = (\frac{x}{r}) * \frac{r^{2}}{y^{2}}[/tex]

    [tex]\therefore x^{2}+y^{2} = ({x}) * \frac{r}{y^{2}}[/tex]

    I am still a little confused on how to get numerical values for x & y. This looks like it can be solved via the unit-circle if I set the right hand side of the equation to 1. Does this look right?

    Thanks~
     
  5. Jun 19, 2013 #4

    LCKurtz

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    That third step is wrong. To divide fractions you "invert and multiply". Also put in ##r^2## on the left side after you fix the right side.
     
  6. Jun 19, 2013 #5

    LCKurtz

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    Also, I missed that you had put ##x^2+y^2## on the left. That is ##r^2##. Your original equation just has ##r## and you should leave it there.
     
  7. Jun 19, 2013 #6
    Yes that was a typo. I actually do the correct method in step 4.
     
  8. Jun 19, 2013 #7
    I am getting the following:
    [tex]r = \frac{x}{r} * \frac{r^{2}}{x^{2}}[/tex]
    [tex]cos \alpha = \frac{x}{r}/[tex]
    [tex]sin \alpha = \frac{y}{r}[/tex]
    [tex]r = \frac{\frac{x}{r}}{\frac{x^{2}}{r^{2}}}[/tex]
    [tex]r = \frac{x}{r} * \frac{r^{2}}{x^{2}}[/tex]
    [tex]r = \frac{x}{r} * \frac{r^{2}}{x^{2}}[/tex]
    [tex]r = \frac{xr}{y^{2}} [/tex]
    [tex]r = \frac{xr}{y^{2}} [/tex]
    [tex]1 = \frac{x}{y^{2}} [/tex]

    Stuck again...now a little confused. Do I need to put this in the form [tex]x^{2} + y^{2} = r^{2}[/tex]
     
  9. Jun 19, 2013 #8

    LCKurtz

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    You want an x-y equation. What's wrong with ##x=y^2##? Now use what you know about xy conics to locate its focus.
     
  10. Jun 19, 2013 #9
    Great. I think I got it. Since this equation is in the form of:
    [tex]x=y^{2}[/tex]
    it is a parabola with the general equation of [tex]x = a(y-k)^{2} + h[/tex]. a = 1 and k = 0 in this case. The focus point for a parabola in this form is at [tex](h+p, k)[/tex] And [tex] p = \frac{1}{4a}[/tex]. Therefore, since a = 1, [tex]p=\frac{1}{4}[/tex].

    The focus point is therefore [tex](\frac{1}{4}, 0)[/tex]

    Thanks LCKurtz!
     
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