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Geometry - Conic Sections

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Hi all,

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

Homework Statement


kxL7ZEC.png



Homework Equations


Ax^2 + Bxy + Cy2 + Dx + Ey + F = 0

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 :)
 

Answers and Replies

  • #2
LCKurtz
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Hi all,

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

Homework Statement


kxL7ZEC.png



Homework Equations


Ax^2 + Bxy + Cy2 + Dx + Ey + F = 0

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 :)
Try putting ##\sin\alpha = y/r## and ##\cos\alpha = x/r## in your equation.
 
  • #3
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Try putting ##\sin\alpha = y/r## and ##\cos\alpha = x/r## in your equation.
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~
 
  • #4
LCKurtz
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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]
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.
 
  • #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.
 
  • #6
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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.
Yes that was a typo. I actually do the correct method in step 4.
 
  • #7
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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.
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]
 
  • #8
LCKurtz
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I am getting the following:

[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]
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.
 
  • #9
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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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