Finding Focus Points of Parabolas in Conic Sections

[skybox]

Hi all,

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

Homework Statement

Homework Equations

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

The Attempt at a Solution

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 :)

 

[LCKurtz]

Hi all,

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

Homework Statement

Homework Equations

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

The Attempt at a Solution

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.

 

[skybox]

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.
x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}

\therefore x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}

\therefore x^{2}+y^{2} = (\frac{x}{r}) * \frac{y^{2}}{r^{2}}

\therefore x^{2}+y^{2} = (\frac{x}{r}) * \frac{r^{2}}{y^{2}}

\therefore x^{2}+y^{2} = ({x}) * \frac{r}{y^{2}}

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~

 

[LCKurtz]

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.
x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}

\therefore x^{2}+y^{2} = \frac{(\frac{x}{r})}{(\frac{y}{r})^{2}}

\therefore x^{2}+y^{2} = (\frac{x}{r}) * \frac{y^{2}}{r^{2}}

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.

 

[LCKurtz]

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.

 

[skybox]

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.

 

[skybox]

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:
r = \frac{x}{r} * \frac{r^{2}}{x^{2}}
cos \alpha = \frac{x}{r}/<br /> sin \alpha = \frac{y}{r}<br /> r = \frac{\frac{x}{r}}{\frac{x^{2}}{r^{2}}}<br /> r = \frac{x}{r} * \frac{r^{2}}{x^{2}}<br /> r = \frac{x}{r} * \frac{r^{2}}{x^{2}}<br /> r = \frac{xr}{y^{2}}<br /> r = \frac{xr}{y^{2}}<br /> 1 = \frac{x}{y^{2}}<br /> <br /> Stuck again...now a little confused. Do I need to put this in the form x^{2} + y^{2} = r^{2}

 

[LCKurtz]

I am getting the following:

1 = \frac{x}{y^{2}}

Stuck again...now a little confused. Do I need to put this in the form x^{2} + y^{2} = r^{2}

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.

 

[skybox]

Great. I think I got it. Since this equation is in the form of:
x=y^{2}
it is a parabola with the general equation of x = a(y-k)^{2} + h. a = 1 and k = 0 in this case. The focus point for a parabola in this form is at (h+p, k) And p = \frac{1}{4a}. Therefore, since a = 1, p=\frac{1}{4}.

The focus point is therefore (\frac{1}{4}, 0)

Thanks LCKurtz!