Parabolas y^2=4px

[Misty]

I'm having trouble with the basic concept here.

I have to find the vertex, focus, and directrix of the given parabola and
sketch its graph.

the first problem is
y^2=8x
The solution manual *magically* transformed this into
4p=8

Could someone explain how y^2=8x gets to 4p=8. Baby steps please.

Thank you,
Misty

[Darryl L. Pierce]

Post by Misty
I'm having trouble with the basic concept here.
I have to find the vertex, focus, and directrix of the given parabola and
sketch its graph.
the first problem is
y^2=8x
The solution manual *magically* transformed this into
4p=8

It's not magic, and it's not the solution. The general form of a parabola is
(y-k)^2=4p(x-h). For your above equation, the parabola is at the origin, so
h and k are both 0. Your p value is within the 8, so the only logical step
is to find your p value by solving the equation:

4p=8

Post by Misty
Could someone explain how y^2=8x gets to 4p=8. Baby steps please.

4p=8 is *not* the solution for y^2=8x. They're trying to show you how to
find the value of *p* with this equation. Then, once you've found your p
value, you plug *that* into the equation y^2=4px.

Hope that helps.

[Darrell]

Post by Darryl L. Pierce
4p=8 is *not* the solution for y^2=8x. They're trying to show you how to
find the value of *p* with this equation. Then, once you've found your p
value, you plug *that* into the equation y^2=4px.

...giving y^2=8x, right back where she started.

Since the task is to determine the vertex, focus, and directrix, not to
mention sketch the graph, the relevence of determining p is to know how far,
and in what direction, the focus is from the vertex, and how far and in what
direction the directrix is from the vertex. IOW it determines where these
are in relation to the vertex. Since p=2, and 2 is positive, this means the
focus lies 2 units to the right of the vertex, i.e. has coordinates (h+2,k),
and the directrix is 2 units on the 'other' side of the vertex, i.e. has
equation x=h-p.

As for why the standard form of a parabola with directrix x=h-p and vertex
(h,k) is
(y-h)^2=4p(x-k), or for that matter for just the cases where the vertex is
at the origin:
y^2=4px
although that was neot specifically asked about, her book should provide
some sort of derivation of that. Actually it would be a good exercise to
try to derive it herself, if she hasn't already done so.

[Misty]

Post by Darryl L. Pierce
It's not magic, and it's not the solution. The general form of a parabola is
(y-k)^2=4p(x-h). For your above equation, the parabola is at the origin, so
h and k are both 0. Your p value is within the 8, so the only logical step
4p=8

Post by Misty
Could someone explain how y^2=8x gets to 4p=8. Baby steps please.

4p=8 is *not* the solution for y^2=8x. They're trying to show you how to
find the value of *p* with this equation. Then, once you've found your p
value, you plug *that* into the equation y^2=4px.

I know its not the solution, it was the first part of the answer in the
solution manual. I just *need* to know how 4p=8 is derived from y^2=8x. If
I can get past that part, then I can do the rest.

Misty

[Darrell]

Post by Misty
I know its not the solution, it was the first part of the answer in the
solution manual. I just *need* to know how 4p=8 is derived from y^2=8x.

If

Post by Misty
I can get past that part, then I can do the rest.

You have been told already by more than one person. Once again, for a
parabola with horizontal axis of symmetry, i.e. opens either rightward or
leftward, the standard form of the equation for vertec (h,k) and directrix
x=h-p is:

(y-h)^2 = 4p(x-h)

By the form listed in the subject line, it appears you are only considering
(at least for now) that the vertex is at the origin, i.e. h and k are both
0. If you take the standard form I just listed and let h and k be 0 you get
exaqctly what you have in the subject line:

y^2 = 4px

Your specific equation is y^2 = 8x. Can't you see this is the same as the
equation right above, with the exception that 4p=8? Can you then solve 4p=8
for p? You have already been told, again by more than one person, what the
significance of p is, so go back and reread those posts if necessary which
tells you exactly what you must do to completely answer the question.

[Darrell]

Post by Darrell

Post by Misty
I know its not the solution, it was the first part of the answer in the
solution manual. I just *need* to know how 4p=8 is derived from y^2=8x.

If

Post by Misty
I can get past that part, then I can do the rest.

You have been told already by more than one person. Once again, for a
parabola with horizontal axis of symmetry, i.e. opens either rightward or
leftward, the standard form of the equation for vertec (h,k) and directrix
(y-h)^2 = 4p(x-h)

a typo....that last h should of course be k.

[Misty]

FWIW, Darrell,

when I replied, there had only been *one* reply and it didn't clarify what I
was trying to figure out. Perhaps I didn't properly articulate my query.
What I couldn't understand was where 4 and P came from. I have since
already figured out what you so condescendingly pointed out.

I was trying to say that my book jumped right to 4p=8 without explaining how
it got to that point. Again, I have figured out that part, though I'm still
confused by the several examples of the distance formula used to prove
y^2=4px

I think the confusion stems from my book using P in addition to p in the
example given along with a graph that points out different (x, y)'s without
clarifying which ones to apply to the formula. In addition, it looks
different from the one the instructor gave us in class last week and I'm
unable to tell if the formula he gave us is essentially the same thing or if
it is to determine something else.

I appreciate all the help I get here, but I think your rudeness was
unnecessary. I suppose you have seen these questions over and over and
maybe for you the answer is extremely obvious. For others, it is not.

Misty

Post by Darrell
You have been told already by more than one person. Once again, for a
parabola with horizontal axis of symmetry, i.e. opens either rightward or
leftward, the standard form of the equation for vertec (h,k) and directrix
(y-h)^2 = 4p(x-h)
By the form listed in the subject line, it appears you are only considering
(at least for now) that the vertex is at the origin, i.e. h and k are both
0. If you take the standard form I just listed and let h and k be 0 you get
y^2 = 4px
Your specific equation is y^2 = 8x. Can't you see this is the same as the
equation right above, with the exception that 4p=8? Can you then solve 4p=8
for p?

Is this kind of sarcasm necessary in a .help forum. Do you now feel more
superior for asking that? You forgot to ask if I can add 2 and 2.

You have already been told, again by more than one person, what the

Post by Darrell
significance of p is, so go back and reread those posts if necessary which
tells you exactly what you must do to completely answer the question.
--
Darrell

[Darrell]

Post by Misty
FWIW, Darrell,
when I replied, there had only been *one* reply and it didn't clarify what I
was trying to figure out.

Sorry, but I don't run Usenet much less control when articles are propogated
from server to server. Suffice it to say you have received several (all
quite helpful, IMO) responses to your question, even if some of them did not
reach you as soon as you would have liked.

<...>

Post by Misty
Perhaps I didn't properly articulate my query.
What I couldn't understand was where 4 and P came from.

I'm not trying to be condescending (really!) but let me mention that I have
already addressed that to some degree. I suggested your book probably has
the derivation, and even suggested that it would be a good exercise for you
to try and derive it yourself. In any event, here's a derivation of y^2=4px
that I hope makes sense.

Start with the definition. A parabola is the set of all points in a plane
that are equidistant from a fixed line, called the directrix, and a fixed
point that's not on the line, called the focus. For now concentrate only on
left or right opening parabolas with vertex at (0,0). Let the directed
distance *from* the vertex *to* the focus be p.

Make a rough sketch of a rightward opening parabola with vertex at the
origin, to include the directrix and focus. You should see how the
coordinates of the focus are (p,0) and the equation of the directrix is
x=-p.

Let (x,y) be any point on the parabola. By the definition, the distance
from any point on the parabola (x,y) to the directrix x=-p and the focus
(p,0) are the same. Use that to write the equation:

dist. from point on parabola to focus
=
dist. from same point on parabola to directrix.

Since the directrix has equation x= -p, the x-coordinate of a point on the
directrix
is "-p"

dist. from (x,y) to (p,0) = dist. from (x,y) to (-p,y)

sqrt[(x-p)^2 + (y-0)^2] = sqrt[(x-(-p))^2 + (y-y)^2]

sqrt[(x-p)^2 + y^2] = sqrt[(x+p)^2 + 0]

sqrt[(x-p)^2 + y^2] = sqrt[(x+p)^2]

Square both sides to remove the sqrt's....

(x-p)^2 + y^2 = (x+p)^2

Expand...

x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2

Collect like terms, geting y^2 alone on one side...

y^2 = x^2 + 2px + p^2 - x^2 + 2px - p^2

y^2 = 4px

The reason it's good to express it this way, is if you are given an equation
either already written in this form (y^2=ax for some "a") or can be
rewritten in this form, you can immediately equate "4p" with "a", and once
*that* equation is solved for p (in your case "a" was 8 giving 4p=8) then it
makes answering the question easy.

vertex: (0,0) ...assumed in the derivation
focus: (p,0)
directrix x= -p

Plug and chug...

focus: (2,0)
directrix: x= -2

It makes sketching a graph pretty easy too, since you have identified these
key features of the graph. As others suggested, you could also detrermine
the latus rectum which in turn tells you additional points on the parabola.

I'll leave it up to you to derive the equation for any general vertex (h,k)
and for that matter, for cases of vertical axis of symmetry (opening
up/down).

<...>

Post by Misty
I appreciate all the help I get here, <...>

You have a strange way of showing it.

[Misty]

I'm getting it now. I've gotten past the 4p thing enough to do the work.

Could you help me apply the distance formula from which x^2=4py is derived,
so that I can understand what I'm really doing. My book's example is not
clear. It has PF=PP' (what is the apostrophe?)
The example graph shows the Focus F(0,p) with an upward parabola with vertex
at the origin and directrix y = -p. They have denoted a point on the
parabola as P(x,y) and a point in the directrix [parrallel to P(x,y)]
denoted by P'(x, -p). So I'm confused. Am I to assume that P and P' and
are the same as p?

So the formula they have says
sqrt[(x-0)^2 + (y-p)^2] = sqrt[(x-x)^2 + (y + p)^2]

OK, I know how to work it out: square both sides, expand, combine and come
up with x^2 = 4py. That I can do.

What has me confused is which p (P, P', p) am I putting in the formula and
which x. It doesn't specify say x1, x2 or whether it is the same x (I know
that sounds stupid, but with the vertex at (0,0) in this example, zeros all
over the place, it appears possible--at least to me since I'm not getting
the mental visual of what I'm doing).

Thank you for your help.

Post by Darryl L. Pierce

Post by Misty
I'm having trouble with the basic concept here.
I have to find the vertex, focus, and directrix of the given parabola and
sketch its graph.
the first problem is
y^2=8x
The solution manual *magically* transformed this into
4p=8

It's not magic, and it's not the solution. The general form of a parabola is
(y-k)^2=4p(x-h). For your above equation, the parabola is at the origin, so
h and k are both 0. Your p value is within the 8, so the only logical step
4p=8

Post by Misty
Could someone explain how y^2=8x gets to 4p=8. Baby steps please.

4p=8 is *not* the solution for y^2=8x. They're trying to show you how to
find the value of *p* with this equation. Then, once you've found your p
value, you plug *that* into the equation y^2=4px.
Hope that helps.
--
Visit the Infobahn Offramp - <http://mypage.org/mcpierce>
"What do you care what other people think, Mr. Feynman?"

[Darryl L. Pierce]

Post by Misty
So the formula they have says
sqrt[(x-0)^2 + (y-p)^2] = sqrt[(x-x)^2 + (y + p)^2]

Better known as the Pythagorean Theorem: r^2 = o^2 + a^2, where r is the
hypotenuese, o is the opposite side and a is the adjacent side.

Post by Misty
OK, I know how to work it out: square both sides, expand, combine and come
up with x^2 = 4py. That I can do.
What has me confused is which p (P, P', p) am I putting in the formula and
which x.

The x is the x coordinate for any point *on* the parabola. The p is the y
coordinate for the focus.

<snip>

Post by Misty
Thank you for your help.

NP. Glad to be of some help. :)

[Greg]

Post by Misty
I'm having trouble with the basic concept here.
I have to find the vertex, focus, and directrix of the given parabola and
sketch its graph.
the first problem is
y^2=8x
The solution manual *magically* transformed this into
4p=8
Could someone explain how y^2=8x gets to 4p=8. Baby steps please.
Thank you,
Misty

Hi Misty,

This is in addition to Darryl's comments (with which I agree).

If a parabola is in the form, y^2 = 4px, then;

1. the vertex is (0,0)
2. the focus is (p,0)
3. the directrix is the line, x = -p and
4. the length of the semi latus rectum (l) is given by, l = 2p.

This is the reason you want to know the value of p.

An aside: In my old 'Maxwell' printed in 1959, it is y^2 = 4ax. Now we have
y^2 = 4px. Is this an algebraic form of inflation?


--
Greg

(remove BALL from address to reply)

[Ian Hutcheson]

General form y^2 = 4px

example given y^2 = 8x

by comparison of the things in front of x,

4p = 8

This tells you that the useful value p in the general case, is 2 in the
example.

Example parabola is y^2 = 8x, so from Gregs posting you now have

1. the vertex is (0,0)

2. the focus is (2,0)

3. the directrix is the line, x = -2 and

4. the length of the semi latus rectum (l) is given by, l = 4

Post again if not clear. Hope this helps,

Ian Hutcheson

[Misty]

Thanks Ian :)

I was later able to see that, but I couldn't at first. Also, "semi latus
rectum" is a term I don't recognize. Is that the same as the primary focal
chord?

Thank you for your help,
Misty

Post by Ian Hutcheson
General form y^2 = 4px
example given y^2 = 8x
by comparison of the things in front of x,
4p = 8
This tells you that the useful value p in the general case, is 2 in the
example.
Example parabola is y^2 = 8x, so from Gregs posting you now have
1. the vertex is (0,0)
2. the focus is (2,0)
3. the directrix is the line, x = -2 and
4. the length of the semi latus rectum (l) is given by, l = 4
Post again if not clear. Hope this helps,
Ian Hutcheson

[Greg]

Post by Misty
Thanks Ian :)
I was later able to see that, but I couldn't at first. Also, "semi latus
rectum" is a term I don't recognize. Is that the same as the primary focal
chord?
Thank you for your help,
Misty

Hi Misty,

I couldn't resist including the 'semi latus rectum'. The latus rectum of a
parabola is the chord through the focus perpendicular to the axis.

In another post you are asking about p, P and P'. I have to guess what your
book says. The lower case p is a parameter to generalise a family of
parabolas, y^2 = 4px. P is an arbitrary point on the parabola P(x_1,y_1)
and P' is the other end of the chord PS where S is the focus.

This may help with your derivation of y^2 = 4px.
Consider the line (the directrix) x = -p and the point S (the focus) with
co-ordinates (p,0).

Then the parabola is the locus of all points, Q(x,y) equidistant from the
directrix and S.

For any Q,
(QS)^2 = y^2 + (p - x)^2 (Pythagoras) [1]
A diagram will help.

The distance Q is from the directrix is (x + p) [2]

This distance squared equals (QS)^2. So, putting [1] and [2] together,
y^2 + (p-x)^2 = (x + p)^2

Expanding and simplifying,
y^2 = 4px.

Hence the locus of Q is y^2 = 4px.

Converse argument gives the position of the directrix and focus of a
parabola expressed in this form.

Hope this helps.


--
Greg

(remove BALL from address to reply)

[Darryl L. Pierce]

Post by Misty
Thanks Ian :)
I was later able to see that, but I couldn't at first. Also, "semi latus
rectum" is a term I don't recognize. Is that the same as the primary
focal chord?

The latus rectum is the line segment that runs through the focus, parallel
to the directrix, that has endpoints on the parabola. The length of the
latus rectum is the focal diameter of the parabola.