Locus HELP (1 Viewer)

epicFAILx · Jul 30, 2011

After missing one lesson of tutoring i am now completely lost.

Can someone please explain to me what a locus is and what its used for!

SpiralFlex · Jul 30, 2011

epicFAILx said:
After missing one lesson of tutoring i am now completely lost.

Can someone please explain to me what a locus is and what its used for!

THE LOCUS

Example 1.

The locus, no it is not an insect. Basically you are given some conditions. For example,

A point $(x, y)$ is always 3 units from the y axis. Find the locus.

The locus is the path or equation a point can take that satisfies the condition set.

In this case, the answer is $x=3$ or $x=-3$ .

See how its perpendicular distance is always 3 units from the $y$ axis as you move along that path?

So the locus is just an equation that satisfies a specific set of conditions/path.

Try these:

1. Find the locus of a point (x, y) that is always 6 units from the y axis.

2. Find the locus of a point (x, y) that is always 4 units from the x axis.

Example 2.

If we were given some harder questions such as,

Find the locus of a point that moves in a plane such that $PA = PB$ . If $A (0, 3)$ and $B = (0, 7)$ .

Woah! What?

Let's us comprehend this "seemingly tricky" phrase. It's basically asking for the path which P can travel such that as it travels, the distance $PA = PB$ . Still remember the distance formula?

$D = \sqrt{(x2-x1)^2+(y2-y1)^2}$

First of all we need to find the distance of $P(x, y)$ to $A(0, 3)$ .

Use the formula,

$PA= \sqrt{(x-0)^2+(y-3)^2}$

$\therefore PA= \sqrt{x^2+(y-3)^2}$

Now, let's find the distance $P(x, y)$ to $B(0, 7)$ .

$PB = \sqrt{(x-0)^2+(y-7)^2}$

$\therefore PB = \sqrt{x^2+(y-7)^2}$

We must now equate them. Since they are the same distance from $P$ .

$PA = PB$

$\sqrt{x^2+(y-3)^2} = \sqrt{x^2+(y-7)^2}$

$x^2+(y-3)^2 = x^2+(y-7)^2$

$y^2 - 6y + 9 = y^2 - 14y + 49$

$8y=40$

$y=8$ [That's your locus.]

Spiral tip: Draw these equations out for all of your graphs, including your locus! You will see what is happening.

Try these:

Find the locus of a point P(x, y) in a plane such that PA = PB:

1. If A = (-2, 3), B = (4, 5)

2. If A = (1, 7), B = (4, -2)

3.

EDITING

epicFAILx · Jul 30, 2011

Ahh, K.

Cool Thanks

But my main problem is with Parabolas. I mean stuff like: vertex, focus, directrix and axis of symettry

SpiralFlex · Jul 30, 2011

epicFAILx said:
Ahh, K.

Cool Thanks

But my main problem is with Parabolas. I mean stuff like: vertex, focus, directrix and axis of symettry

Okay, I shall write a paragraph about that.

SpiralFlex · Jul 30, 2011

THE ANATOMY OF A PARABOLA

http://upload.wikimedia.org/wikipedia/commons/6/66/Parts_of_a_Parabola.JPG

Now. During Year 10 Mathematics when you are a baby your teacher told you there were three parts to a parabola right? The x intercept(s), the y intercept, the vertex and that was it. Now in Year 11 we can extend the definition of the parabola, we give it three more main parts, the focus, the directrix and the latus rectum.

Now you are use to the parabola as a smiley face and a sad face. We can extend the shape of the parabola into two more shapes, the sideways parabola. Kind of like a normal parabola but rotated to the left or right.

Okay back to business!

Your parabola equation can be written in two forms, the general equation form and the vertex form.

$ax^2 + bx + c = 0$ [Where $a \neq 0$ ]

$a(x-h)^2 + k$ [Where $(h,k)$ is our vertex and $a \neq 0$ ]

Mr parabola is very cruel now days. They now expect you to learn the basic form of,

TYPE 1: NORMAL CONCAVE UP PARABOLA.

$(x-h)^2 =4a(y-k)$

Again we got (h, k) as our vertex.

But we got some new visitors! Mr Focal length, Mr Directrix. Better not confuse you, it's called the focal length and the directrix.

The focal length is $a$ and the directrix is I will illustrate this.

I won't introduced the latus rectum just yet.

Now the only way we can comprehend this is to practice.

Example 1.

In the parabola $(x-2)^2=16(y+3)$ . Find:

a. The vertex: Easy, this can be read off it's $(2, -3)$

b. The focus: This is also easy. Recall the form $(x-h)^2 =4a(y-k)$ and notice that our equation resembles that form.

Match everything up. We need to find $a$ . Since that is our focal length.

$4a = 16$

$a = 4$

THAT'S OUR LENGTH, NOT OUR FOCUS POINT!

The focus if you looked at the picture, it's above the vertex. In this case, the focal length was 4 units. It is 4 units above the vertex. Draw the vertex of (2, -3)

We can see the focus is 4 units above it. Hence (2, 1) is the focus point.

c. The directrix was

So the directrix is below the vertex. It maintains 4 units below the vertex. Hence the directrix is (2, -7). However the directrix is an equation of the line.

So $y=-7$

TYPE 2: NORMAL CONCAVE DOWN PARABOLA.

It is in the form of,

$(x-h)^2=-4a(y-k)$

Example 2.

In the parabola $(x+4)^2=-16(y+3)$ . Find:

a. The vertex: Read off, (-4, -3).

b. The focus: $(x+4)^2=-16(y+3)$ is identical to $(x-h)^2=-4a(y-k)$ .

So the focal is,

$-4a = 16$

$a = 4$ [Since it is the focal length.]

Now,

Our vertex (-4, -3). So, 4 below (Since focus is inside parabola and it is concave down.) is the focus.

Our focus is (-4, -7).

c. The directrix: 4 places above your parabola. It's the line $y = 1$

TYPE 3: SIDEWAYS PARABOLA SMILING TO THE RIGHT.

It is in the form of,

$(y-k)^2=4a(x-h)$

Let's see an example.

Example 3.

Can you imagine me smiling to the right of the page? Oh wait. I don't smile. Anyway this parabola is sideways to the right.

If the parabola has an equation of $(y-1)^2=2(x+\frac{5}{2})$ . Find the:

a. Vertex: Read off again! This time be careful. It's $(-\frac{5}{2}, 1)$

b. Focus: $(y-1)^2=2(x+\frac{5}{2})$ is identical to the form $(y-k)^2=4a(x-h)$

The focus is $4a = 2$

$a=\frac{1}{2}$

So if we grab our vertex again

It's $(-\frac{5}{2}, 1)$ , the focus is inside the heart of the parabola. So we can see the heart of the parabola is to the right. If we go two places to the right,

We get that the focus is $(-\frac{3}{2}, 1)$

c. Directrix: The directrix is outside the parabola. The focal length is $\frac{1}{2}$ So two spaces to the left, $x= -\frac{7}{2}$

TYPE 4: SIDEWAYS PARABOLA SMILING TO THE LEFT.

It is in the form of,

$(y-k)^2=-4a(x-h)$

Example 4.

If the parabola has an equation of $(y+1)^2=-2(x+2)$ . Find the:

a. Vertex: Read off again, be careful. It's $(-2, -1)$

b. Focus: We can see that $(y+1)^2=-2(x+2)$ is identical to $(y-k)^2=-4a(x-h)$

So $4a = -2$

$a = \frac{1}{2}$ [Since lengths cannot be negative.]

Our vertex was $(-2, -1)$ . The focus is in the heart of the parabola. Which is to the left. So $\frac{1}{2}$ units to the left of $(-2, -1)$ . That is,

$(-\frac{5}{2}, -1)$

c. Directrix: Outside the heart of the parabola, so $\frac{1}{2}$ to the right. Hence it is $(-\frac{3}{2}, -1)$ . But directrix is an equation. Hence it's $x = -\frac{3}{2}$

Try these:

State the vertex, focus and directrix in these equations and draw the parabola out.

1. $(x-1)^2=4(y+2)$

2. $(x-2)^2 = -4(y+2)$

3. $(y+4)^2 = 4(x+1)$

4. $(y+14)^2 = -4(x+3)$

Enough rambling. Feel free to ask me any questions and I will see if I can help you with understanding the latus rectum.

EDITING: Sorry for being slow, watching Disney's Enchanted.

[Note: Someone should check my working. It's after 9 pm, the time Spiral makes the most mistakes.]

Bored_of_HSC · Jul 30, 2011

SpiralFlex said:
THE LOCUS

Example 1.

The locus, no it is not an insect. Basically you are given some conditions. For example,

A point $(x, y)$ is always 3 units from the y axis. Find the locus.

The locus is the path or equation a point can take that satisfies the condition set.

In this case, the answer is $x=3$ or $x=-3$ .

See how its perpendicular distance is always 3 units from the $y$ axis as you move along that path?

So the locus is just an equation that satisfies a specific set of conditions/path.

Try these:

1. Find the locus of a point (x, y) that is always 6 units from the y axis.

2. Find the locus of a point (x, y) that is always 4 units from the x axis.

Example 2.

If we were given some harder questions such as,

Find the locus of a point that moves in a plane such that $PA = PB$ . If $A (0, 3)$ and $B = (0, 7)$ .

Woah! What?

Let's us comprehend this "seemingly tricky" phrase. It's basically asking for the path which P can travel such that as it travels, the distance $PA = PB$ . Still remember the distance formula?

$D = \sqrt{(x2-x1)^2+(y2-y1)^2}$

First of all we need to find the distance of $P(x, y)$ to $A(0, 3)$ .

Use the formula,

$PA= \sqrt{(x-0)^2+(y-3)^2}$

$\therefore PA= \sqrt{x^2+(y-3)^2}$

Now, let's find the distance $P(x, y)$ to $B(0, 7)$ .

$PB = \sqrt{(x-0)^2+(y-7)^2}$

$\therefore PB = \sqrt{x^2+(y-7)^2}$

We must now equate them. Since they are the same distance from $P$ .

$PA = PB$

$\sqrt{x^2+(y-3)^2} = \sqrt{x^2+(y-7)^2}$

$x^2+(y-3)^2 = x^2+(y-7)^2$

$y^2 - 6y + 9 = y^2 - 14y + 49$

$8y=40$

$y=8$ [That's your locus.]

Spiral tip: Draw these equations all of your graph, including your locus! You will see what is happening.

Try these:

Find the locus of a point P(x, y) in a plane such that PA = PB:

1. If A = (-2, 3), B = (4, 5)

2. If A = (1, 7), B = (4, -2)

3.

EDITING

SpiralFlex said:
THE ANATOMY OF A PARABOLA

http://upload.wikimedia.org/wikipedia/commons/6/66/Parts_of_a_Parabola.JPG

Now. During Year 10 Mathematics when you are a baby your teacher told you there were three parts to a parabola right? The x intercept(s), the y intercept, the vertex and that was it. Now in Year 11 we can extend the definition of the parabola, we give it three more main parts, the focus, the directrix and the latus rectum.

Now you are use to the parabola as a smiley face and a sad face. We can extend the shape of the parabola into two more shapes, the sideways parabola. Kind of like a normal parabola but rotated to the left or right.

Okay back to business!

Your parabola equation can be written in two forms, the general equation form and the vertex form.

$ax^2 + bx + c = 0$ [Where $a \neq 0$ ]

$a(x-h)^2 + k$ [Where $(h,k)$ is our vertex and $a \neq 0$ ]

Mr parabola is very cruel now days. They now expect you to learn the basic form of,

$x^2 =4ay$

EDITING: Sorry for being slow, watching Disney's Enchanted.

I'm dedicating my 50th post to this guy, so i can finally rep him for helping peeps out.

SpiralFlex · Jul 30, 2011

Bored_of_HSC said:
I'm dedicating my 50th post to this guy, so i can finally rep him for helping peeps out.

Thank you human. My head spins around and around. I need to snack on some humans.

SpiralFlex · Jul 30, 2011

Oh no! Did you mean the locus in the path of a parabola not the geometry of a parabola? Silly Spiral.

tambam · Jul 30, 2011

You should just sign up for this online tutoring site, sponsored by maccas so its free.
http://www.mathsonline.com.au/
It's pretty good, covers all topics in depth.
And the guy has a really soothing voice and often tells you you're excellent, so its morale boosting too (Y)

SpiralFlex · Jul 30, 2011

tambam said:
You should just sign up for this online tutoring site, sponsored by maccas so its free.
http://www.mathsonline.com.au/
It's pretty good, covers all topics in depth.
And the guy has a really soothing voice and often tells you you're excellent, so its morale boosting too (Y)

Almost all videos end with "excellent, good luck with your questions". By the way, your rabbit is delicious.

epicFAILx · Jul 31, 2011

tambam said:
You should just sign up for this online tutoring site, sponsored by maccas so its free.
http://www.mathsonline.com.au/
It's pretty good, covers all topics in depth.
And the guy has a really soothing voice and often tells you you're excellent, so its morale boosting too (Y)

I have tutoring already. Isn't one enough?

juicystar07 · Jul 31, 2011

I swear @SpiralFlex, you are amazing. I can't wait until we see the ATAR scores, and we shall see your name next to the top mark, 99.95

SpiralFlex · Jul 31, 2011

juicystar07 said:
I swear @SpiralFlex, you are amazing. I can't wait until we see the ATAR scores, and we shall see your name next to the top mark, 99.95

My name is sacred.

powlmao · Jul 31, 2011

Is this Ext 1??

Alkanes · Jul 31, 2011

powlmao · Jul 31, 2011

fuck, my school is only up tto functions in 2U

Uniqueness · Jul 31, 2011

In example 3 type 3, since $\left ( \textit{y - 1}\right )^{2}= 2 \left ( x + \frac{5}{2} \right )$ shouldn't the vertex be $\left ( -\frac{5}{2}, 1 \right )$ ? Or did I do something wrong again?

Hermes1 · Jul 31, 2011

Uniqueness said:
In example 3 type 3, since $\left ( \textit{y - 1}\right )^{2}= 2 \left ( x + \frac{5}{2} \right )$ shouldn't the vertex be $\left ( -\frac{5}{2}, 1 \right )$ ? Or did I do something wrong again?

yeh ur right. spiral made a mistake. u can hardly blame him after all that working he had to do.

tambam · Jul 31, 2011

epicFAILx said:
I have tutoring already. Isn't one enough?

Haha, no its not actually tutoring, it just has videos/explanations for all the topics covered in all years of maths.
So it would be ideal in a situation like yours where you've missed a lesson on locus, you could just go watch the lesson & print out the summary on the website.
Then spiralflex would not have needed to type up that lengthy, albeit quite excellent, explanation.

& thank you, my bunny is awesome

juicystar07 · Jul 31, 2011

SpiralFlex said:
My name is sacred.

I bow down to you, SpiralFlex. Give me your wisdom! HAHAHAHA

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