# Thread: Locus Help

1. Here is another problem.
Write the equation of the parabola with focus $\displaystyle (-1,1)$ and directrix $\displaystyle y=2x-1$.

2. Originally Posted by Plato
Here is another problem.
Write the equation of the parabola with focus $\displaystyle (-1,1)$ and directrix $\displaystyle y=2x-1$.
Is it OK for anyone to chime in?.

This is a different parabola problem. Cool.

F(-1,1) and directrix is the line y=2x-1.

Standard form of line is $\displaystyle y-2x+1=0$

We can use this in the right side of the following equation, which is the distance form a point to a line formula.

$\displaystyle \sqrt{(x+1)^{2}+(y-1)^{2}}=\frac{|y-2x+1|}{\sqrt{1^{2}+(-2)^{2}}}$

Square both sides and expand:

$\displaystyle x^{2}+2x+y^{2}-2y+2=\frac{4}{5}x^{2}-\frac{4}{5}xy-\frac{4}{5}x+\frac{1}{5}y^{2}+\frac{2}{5}y+\frac{1 }{5}$

$\displaystyle \frac{1}{5}x^{2}+(\frac{4}{5}y+\frac{14}{5})x+\fra c{4}{5}y^{2}-\frac{12}{5}y+\frac{9}{5}=0$

$\displaystyle x^{2}+(4y+14)x+4y^{2}-12y+9=0$

And that is the equation of said parabola. Assuming I didn't make a mistake.

3. Originally Posted by galactus
Is it OK for anyone to chime in?.

This is a different parabola problem. Cool.

F(-1,1) and directrix is the line y=2x-1.

Standard form of line is $\displaystyle y-2x+1=0$

We can use this in the right side of the following equation, which is the distance form a point to a line formula.

$\displaystyle \sqrt{(x+1)^{2}+(y-1)^{2}}=\frac{|y-2x+1|}{\sqrt{1^{2}+(-2)^{2}}}$

Square both sides and expand:

$\displaystyle x^{2}+2x+y^{2}-2y+2=\frac{4}{5}x^{2}-\frac{4}{5}xy-\frac{4}{5}x+\frac{1}{5}y^{2}+\frac{2}{5}y+\frac{1 }{5}$

$\displaystyle \frac{1}{5}x^{2}+(\frac{4}{5}y+\frac{14}{5})x+\fra c{4}{5}y^{2}-\frac{12}{5}y+\frac{9}{5}=0$

And that is the equation of said parabola. Assuming I didn't make a mistake.
Looks good to me!

-Dan

4. WHEW!!. Good. I'm glad that wasn't for nothing.

5. For the hyperbola (okay, I had to look this one up ):

"An" hyperbola is the locus of points where the difference in the distance to two fixed points (foci) is constant.

Going back to the ellipse, I don't know what kind of figure is given if you have, say, three or more foci. Is this even defined?

-Dan

6. Originally Posted by Plato
Here is another problem.
Write the equation of the parabola with focus $\displaystyle (-1,1)$ and directrix $\displaystyle y=2x-1$.
Cool, let me try.

So using Topsquark's method:

$\displaystyle d_1 = \sqrt{x^2 + y^2}$

$\displaystyle d_2 = y - (2x - 1)$

$\displaystyle \sqrt{x^2 + y^2} = y - 2x + 1$

$\displaystyle x^2 + y^2 = (y - 2x + 1)^2$

$\displaystyle x^2 + y^2 = y^2 - 4xy - 4x + 2y + 4x^2 + 1$

$\displaystyle y^2 - y^2 = 3x^2 - 4xy + 2y - 4x + 1$

$\displaystyle -2y = 3x^2 - 4xy - 4x + 1$

$\displaystyle -2y + 4xy = 3x^2 - 4x + 1$

$\displaystyle y(-2 + 4x) = 3x^2 - 4x + 1$

$\displaystyle y = \frac{3x^2}{(-2 + 4x)} - \frac{4x}{(-2 + 4x)} + \frac{1}{(-2 + 4x)}$

Is this right?

7. Thanks a lot for all those definitions Topsquark, it really helps.

8. Originally Posted by janvdl
Cool, let me try.

So using Topsquark's method:

$\displaystyle d_1 = \sqrt{x^2 + y^2}$

$\displaystyle d_2 = y - (2x - 1)$

$\displaystyle \sqrt{x^2 + y^2} = y - 2x + 1$

$\displaystyle x^2 + y^2 = (y - 2x + 1)^2$

$\displaystyle x^2 + y^2 = y^2 - 4xy - 4x + 2y + 4x^2 + 1$

$\displaystyle y^2 - y^2 = 3x^2 - 4xy + 2y - 4x + 1$

$\displaystyle -2y = 3x^2 - 4xy - 4x + 1$

$\displaystyle -2y + 4xy = 3x^2 - 4x + 1$

$\displaystyle y(-2 + 4x) = 3x^2 - 4x + 1$

$\displaystyle y = \frac{3x^2}{(-2 + 4x)} - \frac{4x}{(-2 + 4x)} + \frac{1}{(-2 + 4x)}$

Is this right?
I'm afraid not. galactus got it right in all the details. Plato's problem was (fortunately) not typical!

The method I posted works only for a parabola that opens either upward or downward. Plato's parabola opens off on a diagonal, so this is more complicated.

Try this for a more general problem (again using an upward or downward opening parabola):
Given a focus F(a, b) and a directrix y = c the distance from the point (x, y) to F is:
$\displaystyle d_1 = \sqrt{(x - a)^2 + (y - b)^2}$

and the vertical distance from (x, y) to the line y = c is
$\displaystyle d_2 = y - c$.

You finish it from here.

-Dan

9. Originally Posted by topsquark
Plato's problem was (fortunately) not typical!
I'm only still in school, go easy on me!

10. Originally Posted by janvdl
I'm only still in school, go easy on me!
Aren't we all?

-Dan

11. Originally Posted by topsquark
Aren't we all?

-Dan
The only problem is that most of you guys are at university or already have math degrees. I can't wait for next year so that i can start learning REAL maths.

12. It's obvious you love math, so you're 'shining times' are ahead of you when it comes to learning it. It's amazing how few people enjoy it and most look at it as a disease they have to keep from catching. I can see it now:

Janvdl, PhD.; Fields Medalist; solver of the Riemann Zeta problem.

Actually, I didn't get an interest in math until I started college. I initially majored in Comp Sci, but changed to math once I took calculus.

I had used trig and so forth while surveying, but my interest peaked in school.

Have fun. Oh, BTW, did you follow that unorthodox parabola solution?.

13. Originally Posted by galactus

Janvdl, PhD.; Fields Medalist; solver of the Riemann Zeta problem.
I'd give you 10 thanks for that one if i could
No, i really do love maths and i will be studying B.Com Actuarial Science next year, and after that i will continue studying B.Sc Mathematics and Applied Mathematics until i do reach my PhD in it.

Originally Posted by galactus
I had used trig and so forth while surveying, but my interest peaked in school.
To be honest, school maths bores me. Except for locii, which i still need to get the hang of. But i love calculus, its by far the maths i've had the most fun with. I like the whole concept of number theory as well, but i struggle to understand it, primarily because no-one ever taught me stuff like it.

Originally Posted by galactus
...did you follow that unorthodox parabola solution?
I lost you on the second step i think Where you got an absolute value as a numerator.

And to be honest, i still don't really know what a directrix is. It has something to do with a distance moving away from the parabola or something, but i don't get it.

14. The right side of the equation was the formula for the distance form a point to a line. You can't have a negative distance, so it's an absolute value.

A parabola is defined as a set of points equidistant from a fixed point(focus) and a fixed line(directrix).

Let me show you a graph. Let's just use $\displaystyle y=\frac{-1}{6}x^{2}$.

The distance from the directrix to any point on the parabola is the same as the distance from the focus to that point. That's what a parabola is.

The distance from the focus to the vertex is the same as the distance form the vertex to the directrix.

The distance, normally denoted by 'p', is the distance from the focus to the vertex. $\displaystyle p=\frac{1}{4a}$

In this case, a=-1/6, so $\displaystyle \frac{1}{4(-1/6)}=-3/2$. See?. Does that help a little?.

15. Originally Posted by galactus
The distance from the directrix to any point on the parabola is the same as the distance from the focus to that point. That's what a parabola is.
Am i correct in assuming that the focus and the directrix could be moved like we wish? As long as their distances to a certain point on the parabola remain equal?

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