1. ## Locus Help

We recently started with locus in Maths. Analytical Geometry to be precise, and I while I don't find it too hard, I do have a few questions about it.

1. How do you know when exactly the locus is a circle, or a parabola, or a straight line, etc? Can someone give me a few tips on this?

2. Are there any interesting things we can do with a locus? Or are they just used to find a point that is just as far from point A as point B...

3. What is the plural of locus?

2. Originally Posted by janvdl
We recently started with locus in Maths. Analytical Geometry to be precise, and I while I don't find it too hard, I do have a few questions about it.
You have school in July?

1. How do you know when exactly the locus is a circle, or a parabola, or a straight line, etc? Can someone give me a few tips on this?
With analytic geometry.
Let $(0,0)$ be an origin. Let $(x,y)$ be any point. And we want to find the locus of points equidistant from $(0,0)$ 1 unit. So by the distance formula, $\sqrt{x^2+y^2}=1$ thus, $x^2+y^2=1$. Which is a circle.

3. What is the plural of locus?
Locii

3. Originally Posted by ThePerfectHacker
You have school in July?
Yes we have a very short June holiday. Same with April and September. December holiday is about 2 months.

[tex] With analytic geometry.
Let $(0,0)$ be an origin. Let $(x,y)$ be any point. And we want to find the locus of points equidistant from $(0,0)$ 1 unit. So by the distance formula, $\sqrt{x^2+y^2}=1$ thus, $x^2+y^2=1$. Which is a circle.
How would i know when to use a hyperbole? Or a parabole?

4. Originally Posted by janvdl
How would i know when to use a hyperbole? Or a parabole?
Do you know the definition of what a parabola is in terms of locus?

5. Originally Posted by ThePerfectHacker
Do you know the definition of what a parabola is in terms of locus?
No, they never teach us anything, we have to figure it out ourselves.

6. A parabola is the locus of points in the plane equally distant from a fixed line D called the directrix and a fixed point F, not on D, called the focus.

7. OK, we were given a tutorial about locii and we were taught NOTHING about how to apply it like this... It also involves series and stuff.

Please give explanations as well guys, thanks.

Write 2,45 as a normal fraction, using $S_{\infty}$.
The ,45 repeats itself... Like 2,45454545...

8. Originally Posted by janvdl
OK, we were given a tutorial about locusts and we were taught NOTHING about how to apply it like this...
I should hope not. Locusts are bugs, by the way. I think you mean either loci or locii, depending on who is teaching you.

Originally Posted by janvdl
Write 2,45 as a normal fraction, using $S_{\infty}$.
The ,45 repeats itself... Like 2,45454545...
$2. \bar{45} = 2 + \sum_{n = 1}^{\infty}45 \cdot \left ( \frac{1}{100} \right ) ^n$

The last term is a geometric series. And we know that
$S_{\infty} = \sum_{n = 0}^{\infty} ar^n = \frac{a}{1 - r}$

So
$\sum_{n = 1}^{\infty}45 \cdot \left ( \frac{1}{100} \right ) ^n = \sum_{n = 0}^{\infty}45 \cdot \left ( \frac{1}{100} \right ) ^n - 45 = \frac{45}{1 - \frac{1}{100}} - 45$

$= \frac{45}{\frac{99}{100}} - 45 = 45 \left ( \frac{100}{99} - 1 \right )$

$= 45 \cdot \frac{1}{99} = \frac{45}{99}$

Thus

$2. \bar{45} = 2 + \sum_{n = 1}^{\infty}45 \cdot \left ( \frac{1}{100} \right ) ^n = 2 + \frac{45}{99}$

$= \frac{243}{99}$

-Dan

9. Personally I think the following way is both easier and faster than the series method.

Let S = $2. \bar{45}$

Or
$S = 2.45454545454545 ...$

Then $100S = 245.4545454545 ...$

When we subtract the two:
$100S - S = 245.4545454545 ... - 2.4545454545 ...$

$99S = 243$

$S = \frac{243}{99}$
without having to memorize formulas and stuff.

-Dan

10. Sorry, i meant locii. What was i thinking?

Thanks Topsquark.

Let the focus be at $(0,0)$ and the the directrix be $y=-1$.
Find the equation of the parabola.

12. Originally Posted by ThePerfectHacker

Let the focus be at $(0,0)$ and the the directrix be $y=-1$.
Find the equation of the parabola.
$4p(y - k) = (x - h)^2$

Set $4p = -1$ then $p = - \frac{1}{4}$

$-1(y - 0) = (x - 0)^2$

$y = -x^2$

13. Originally Posted by ThePerfectHacker

Let the focus be at $(0,0)$ and the the directrix be $y=-1$.
Find the equation of the parabola.
Originally Posted by janvdl
$4p(y - k) = (x - h)^2$

Set $4p = -1$ then $p = - \frac{1}{4}$

$-1(y - 0) = (x - 0)^2$

$y = -x^2$
The problem with your method is that $(h, k)$ is the coordinates of the vertex point, not the focus.

Here's the derivation. Perhaps this will be of value to you.
F(0, 0) and directrix y = -1.

We wish to find the locus of points such that they are equidistant from the focus and the line.

Call the general point where this occurs (x, y).

The distance from the focus (0, 0) and the point (x, y) is
$d_1 = \sqrt{x^2 + y^2}$

The (vertical) distance from (x, y) to the line y = -1 is
$d_2 = y - (-1)$

We set these two distances to be equal and this gives us a condition on x and y.

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

Now solve for y:
$x^2 + y^2 = (y + 1)^2$

$x^2 + y^2 = y^2 + 2y + 1$

$x^2 = 2y + 1$

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

-Dan

14. ## .

if point are all equidistant from x, it is a circle. if they are equidistand from x and y, it is a line.

locii is plural of locus.