B What is the focus and parameter of a parabola with vertex off the origin?

[rudransh verma]

TL;DR Summary

Problem in general eqn of parabola

The general eqn of parabola is $(x-h)^2=-4a(y-k)$. This is the parabola whose vertex doesn't lie on origin and axis is parallel to y axis. It opens downwards. Vertex is (h,k). What will be the focus of this parabola and what is $a$ in general form?
In the diagram a<0 which is confusing.https://byjus.com/maths/standard-equations-of-parabola/

 

[fresh_42]

Summary:: Problem in general eqn of parabola

In the diagram a<0 which is confusing.https://byjus.com/maths/standard-equations-of-parabola/

No, it is wrong. Parabolas of the form $y=ax^2 +\ldots$ are open at the top if $a>0$ and open at the bottom if $a<0.$ You can see this by yourself if you plug in some values for the parameters and the variable $x.$

It could, however, also be the case that you put in a minus sign where there is none and the actual equation would be $(x-h)^2=4a(y-k).$

You can find a lot of information on the Wikipedia page:
https://en.wikipedia.org/wiki/Parabola

 

[rudransh verma]

No, it is wrong. Parabolas of the form $y=ax^2 +\ldots$ are open at the top if $a>0$ and open at the bottom if $a<0.$ You can see this by yourself if you plug in some values for the parameters and the variable $x.$

It could, however, also be the case that you put in a minus sign where there is none and the actual equation would be $(x-h)^2=4a(y-k).$

You can find a lot of information on the Wikipedia page:
https://en.wikipedia.org/wiki/Parabola

If I open this up I get $\frac{-x^2}{4a}+\frac{hx}{2a}+\frac{4ak-h^2}{4a}=y$
Where is a>0?

 

[fresh_42]

If I open this up I get $\frac{-x^2}{4a}+\frac{xh}{2a}+\frac{4ak-h^2}{4a}=y$

What is this? If $a<0$ the coefficient at $x^2$ is positive, i.e. open at the top.

 

[Mark44]

Summary:: Problem in general eqn of parabola

The general eqn of parabola is $(x-h)^2=-4a(y-k)$. This is the parabola whose vertex doesn't lie on origin and axis is parallel to y axis. It opens downwards. Vertex is (h,k). What will be the focus of this parabola and what is $a$ in general form?
In the diagram a<0 which is confusing.https://byjus.com/maths/standard-equations-of-parabola/

The equation above is not the "general equation of a parabola." The page you linked to shows 8 different forms of the equation, depending on whether the vertex is at the origin (the first four) or where the vertex has been translated away from the origin.

The four equations in which the parabola opens to the left or right don't represent functions, so you're probably better off focusing on the ones that open upward or downward.

 

[phinds]

whose vertex doesn't lie on origin

Aside from all other considerations about your post, you need to be aware that this is a science forum and precision of statements is important. That statement is NOT a generally true statement because it is false if h and k are zero. The general statement would be "whose vertex is at (h,k)".

 

[rudransh verma]

The equation above is not the "general equation of a parabola."

Then what is the general eqn of parabola? Is it $y=ax^2+bx+c$

 

[Mark44]

Then what is the general eqn of parabola? Is it $y=ax^2+bx+c$

This is the general equation of a parabola that opens upward (a > 0) or downward (a < 0). It is not the equation of a parabola that opens either to the left or the right, so this equation is not the general equation of a parabola.

The equation you wrote is general (in the sense of upward-opening or downward-opening parabolas), but is not helpful at all in determining the vertex of a parabola.

 

[FactChecker]

Summary:: Problem in general eqn of parabola

The general eqn of parabola is $(x-h)^2=-4a(y-k)$. This is the parabola whose vertex doesn't lie on origin and axis is parallel to y axis. It opens downwards. Vertex is (h,k). What will be the focus of this parabola and what is $a$ in general form?
In the diagram a<0 which is confusing.https://byjus.com/maths/standard-equations-of-parabola/

I disagree with the diagram. If $y \lt k$, then $(y-k) \lt 0$. So if $a \lt 0$ and $y \lt k$, the right-hand side, $-4a(y-k)$, is negative. But the left-hand side, $(x-h)^2$, is always positive.

 

[rudransh verma]

The page you linked to shows 8 different forms of the equation,

You mean 8 forms of the general eqn of parabola which is?

 

[Mark44]

You mean 8 forms of the general eqn of parabola which is?

There is no "general equation of a parabola." Period. If the parabola opens upward or downward, its equation is very different from one that opens to the left or to the right.

In my earlier post I said that the page you linked to showed 8 different equations, each of which shows one type of parabola. Again, the first four equations show parabolas whose vertices are at the origin, and the next four equations show parabolas whose vertices are at a point (h, k).

Notice that the title of the page you linked to is "Standard Equations of Parabola (sic)".

 

[rudransh verma]

There is no "general equation of a parabola." Period. If the parabola opens upward or downward, its equation is very different from one that opens to the left or to the right.

But i don't understand where point (a,0) or (-a,0) or (0,a)... lies in last four eqns on the graph.

 

[phinds]

But i don't understand where point (a,0) or (-a,0) or (0,a)... lies in last four eqns on the graph.

Your confusion clearly stems from two things. First you don't actually understand the relationship between the equations and the graphs and second you clearly have not actually done the work to graph some of the equations yourself. The first problem stems directly from the second problem.

You say you don't know where "a" is. That's because "a" is a variable, as you would have immediately realized had you graphed one of the equations. If you make "a" big vs little, you get a signification different looking graph (but all still parabolas, just some are "wide" and some are "narrow" depending on what value you give to "a").

As usual, you don't seem to be taking a "bottom's up" approach to learning but rather insist on jumping into the middle of things before actually understanding the basics. It's an extraordinarily poor learning technique and will continue to serve you poorly.

 

[jbriggs444]

You say you don't know where "a" is. That's because "a" is a variable

$a$ is a variable, yes. One might call it a "parameter" that describes the parabola. It is the distance between the "vertex" of the parabola and its "focus".

A parabola can be described based on a given line (the "directrix") and a given point (the "focus"). The point is called the "focus". The parabola is the set of points that are at the same distance from the directrix as they are from the focus.

For the parabola described by the equation: y = x^2, one can guess that the focus is at $(0,\frac{1}{4}$) and that the directrix is at $y=-\frac{1}{4}$. We can test this with a few values from the parabola.

$(x,y) = (0,0)$: Distance from focus = $\frac{1}{4}$, Distance from directrix = $\frac{1}{4}$.

$(x,y) = (\frac{1}{2},\frac{1}{4}$): Distance from focus: $\frac{1}{2}$, Distance from directrix: $\frac{1}{2}$.

$(x,y) = (1,1)$: Distance from focus = $\frac{5}{4}$, Distance from directrix = $\frac{5}{4}$.

$(x,y) = (2,4)$: Distance from focus = $\frac{17}{4}$, Distance from directrix = $\frac{17}{4}$.

Interesting way to generate Pythagorean triples.

 

[rudransh verma]

la. It is the distance between the "vertex" of the parabola and its "focus".

Ok! But if it’s the distance how come it’s negative or positive?

If you make "a" big vs little, you get a signification different looking graph (but all still parabolas, just some are "wide" and some are "narrow" depending on what value you give to "a").

I have checked the graphs already. But the confusion is this that I mentioned above.

 

[mcastillo356]

Ok! But if it’s the distance how come it’s negative or positive?

I have checked the graphs already. But the confusion is this that I mentioned above.

Hi, @rudransh verma, I'm not native, could you explain me the first sentence of the quote?
Love

 

[Lnewqban]

 

[phinds]

View attachment 298850

Nice. Too bad that the OP is either unable or unwilling to do that for himself, as I had suggested to him.

 

[FactChecker]

The root problem is with the diagram shown in the OP, which contradicts itself. It says that $a$ is negative but then has a graph that does not match.

 

[phinds]

The root problem is with the diagram shown in the OP

No, I think the root problem is that the OP has not tried to graph parabolas himself, using various versions of the equation/formula, to gain an understanding of the physical meaning of the various elements in the equations.

 

[FactChecker]

No, I think the root problem is that the OP has not tried to graph parabolas himself,

The equation given, with the statement in the diagram that $a$ is negative, does not match the graph. That is a problem.

 

[symbolipoint]

Discussion looks too wild and I cannot follow it too closely. At least someone discussed the use of the distance formula definition for a parabola. That coordinated from the proper picture/sketch should be extremely helpful. The rest is to continue study, exercise, and think.

 

[FactChecker]

Discussion looks too wild and I cannot follow it too closely. At least someone discussed the use of the distance formula definition for a parabola. That coordinated from the proper picture/sketch should be extremely helpful. The rest is to continue study, exercise, and think.

This discussion has wandered from the original OP question. The original question is about the bottom diagram in the linked article. That diagram is wrong. A negative $a$ in the equation shown can not give the graph shown.

 

[rudransh verma]

This discussion has wandered from the original OP question. The original question is about the bottom diagram in the linked article. That diagram is wrong. A negative $a$ in the equation shown can not give the graph shown.

Yes! I think it’s wrong too. But how come a website like this has mistakes. It’s quite popular in India now a days.
It should open up.

 

[FactChecker]

Yes! I think it’s wrong too. But how come a website like this has mistakes. It’s quite popular in India now a days.
It should open up.

Proofreading is a never-ending, thankless job. No matter how long and hard they work, some errors will remain. You did well to spot the error.

 

[rudransh verma]

Proofreading is a never-ending, thankless job. No matter how long and hard they work, some errors will remain. You did well to spot the error.

Ok. Thank you.

 

[fresh_42]

But how come a website like this has mistakes. It’s quite popular in India now a days.

As I said: Maybe the equation actually was $(x-h)^2=+4a(y-k).$ (The link does not open here.)

 

[FactChecker]

As I said: Maybe the equation actually was $(x-h)^2=+4a(y-k).$ (The link does not open here.)

Yes, that would fix the figure in the linked article. Probably better would be if it said $a \gt 0$ to conform to the interpretation of $a$ as a distance. But the figure is wrong as it stands. Some other figures that say $a \lt 0$ are wrong.
(The link works in my Firefox browser)

 

[fresh_42]

(The link works in my Firefox browser)

You don't have the BCP handicap (Brussel's chair pooper = EU).

 

[neilparker62]

https://www.desmos.com/calculator/ykn3kla3ta

Here you can experiment with varying the parameters in your equation.

 

[symbolipoint]

Proofreading is a never-ending, thankless job. No matter how long and hard they work, some errors will remain. You did well to spot the error.

Mistakes happen, and a few of them are published. A student on some rare occasion will ask for help somewhere, tell his audience what "the book" or "the answer key" says, and upon some helper checking on the problem, finds that the book or key answer is wrong. Yes! Mistakes in answer key publishing happen!

 

[neilparker62]

No, it is wrong. Parabolas of the form $y=ax^2 +\ldots$ are open at the top if $a>0$ and open at the bottom if $a<0.$ You can see this by yourself if you plug in some values for the parameters and the variable $x.$

It could, however, also be the case that you put in a minus sign where there is none and the actual equation would be $(x-h)^2=4a(y-k).$

You can find a lot of information on the Wikipedia page:
https://en.wikipedia.org/wiki/Parabola

Yes - if you include a minus sign in the "general" equation then when a is positive the graph opens downward which is incorrect. This is like textbooks which insist on writing the equation of a parabola in turning point form as $$y=a(x+p)^2+q$$ instead of $$y=a(x-p)^2+q$$. We should be using the latter so that a positive value of p results in a right shift.

 

[symbolipoint]

Yes - if you include a minus sign in the "general" equation then when a is positive the graph opens downward which is incorrect. This is like textbooks which insist on writing the equation of a parabola in turning point form as $$y=a(x+p)^2+q$$ instead of $$y=a(x-p)^2+q$$. We should be using the latter so that a positive value of p results in a right shift.

If that "a" is positive value, then the parabola has a minimum point. If "a" is negative, then parabola has a maximum point.

 

[neilparker62]

If that "a" is positive value, then the parabola has a minimum point. If "a" is negative, then parabola has a maximum point.

Yes - that's correct and that's how it should be. eg $y=x^2$ with a=1 !

 

[robphy]

https://www.desmos.com/calculator/ykn3kla3ta

Here you can experiment with varying the parameters in your equation.

Here's a tweak with some fancy Desmos tricks.
https://www.desmos.com/calculator/awuq5gbfvk
[long click the control circles to see]

 

[rudransh verma]

Here's a tweak with some fancy Desmos tricks.
https://www.desmos.com/calculator/awuq5gbfvk
[long click the control circles to see]

First I want to clear one thing. If its established that a is a distance and cannot be negative or positive like you took in desmos graph then why are we talking about a<0 or a>0 in some of the posts and making graphs of it too?

 

[jbriggs444]

First I want to clear one thing. If its established that a is a distance and cannot be negative or positive like you took in desmos graph then why are we talking about a<0 or a>0 in some of the posts and making graphs of it too?

In the case of a parabola that is oriented vertically (opens either up or down, not sideways or diagonally), $a$ can be treated as a displacement rather than as a distance.

 

[rudransh verma]

In the case of a parabola that is oriented vertically (opens either up or down, not sideways or diagonally), $a$ can be treated as a displacement rather than as a distance.

I think to avoid mistakes I should memorize the first four equations of parabola ie $y^2=4ax, y^2=-4ax, x^2=4ay, x^2=-4ay$
Other than that all other equations involving h,k just are the extensions of these four equations whose vertices don't lie on origin. No matter what the equation of the parabola is just convert it to these eight eqns and the sign before a will tell the direction where the parabola opens up.
$y^2 = 4ax$ Taking a>0, opens right. Taking a<0, opens left.
$y^2 = -4ax$ Taking a>0, opens left. Taking a<0, opens right.
$x^2 = 4ay$ Taking a>0, opens up. Taking a<0, opens down.
$x^2 = -4ay$ Taking a>0, opens down. Taking a<0, opens up.
$(y – k)^2 = 4a(x – h)$ Taking a>0, opens right. Taking a<0, opens left.
$(y – k)^2 = -4a(x – h)$ Taking a>0, opens left. Taking a<0, opens right.
$(x – h)^2 = 4a(y – k)$ Taking a>0, opens up. Taking a<0, opens down.
$(x – h)^2 = -4a(y – k)$ Taking a>0, opens down. Taking a<0, opens up.

 

[robphy]

$y^2 = 4ax$ Taking a>0, opens right. Taking a<0, opens left.
$y^2 = -4ax$ Taking a>0, opens left. Taking a<0, opens right.

Of course the second equation is equivalent to the first equation,
and may be more clear if the pair is written as
$y^2 = 4(\ a\ )x$ Taking a>0, opens right. Taking a<0, opens left.
$y^2 = 4(-a)x$ Taking a>0, opens left. Taking a<0, opens right.

or

$y^2 = 4(\ a\ )x$ Taking a>0, opens right. Taking a<0, opens left.
$y^2 = 4(-a)x$ Taking (-a)>0, opens right. Taking (-a)<0, opens left.

But if it helps you do the bookkeeping, stick with it [for now].

 

[neilparker62]

Graph including focus and directrix and an attempt to show distance of (x,y) from both is the same.

https://www.desmos.com/calculator/17xqfbn4tp

Edit: added a circle centred at a point on the parabola and showing distance to focal point is the same as distance to directrix.

 

[rudransh verma]

In the case of a parabola that is oriented vertically (opens either up or down, not sideways or diagonally), $a$ can be treated as a displacement rather than as a distance.

You mean in physics. The correct way should be to fix $a>0$ that resembles the definition and use different versions of eqn to plot the graph. $y^2=4ax$ taking $a>0$ and $y^2=-4ax$ taking $a<0$ are basically the same eqns.

By the way can you convert $y=\tan \theta x- \frac {gx^2}{2(v\cos \theta)^2}$ into standard form like $x^2=4ay$
I am unable to do it. This eqn is in the form $y=ax^2+bx+c$
If it can’t be done then how are two standard forms related (like x^2=4ay and y=ax^2+bx+c).

 

[rudransh verma]

Graph including focus and directrix and an attempt to show distance of (x,y) from both is the same.

Nice effort but two a's in the first eqn? Is it correct? I don't think its in most simplified form.

 

[neilparker62]

Nice effort but two a's in the first eqn? Is it correct? I don't think its in most simplified form.

It would not work if not correct. However I have modified to use the correct variables: $$(x-h)^2=4a(y-(d+a))$$ https://www.desmos.com/calculator/17xqfbn4tp

A glance at the parabola should tell you that 'k' the vertical shift parameter in your equation equals d+a where y=d is the equation of the directrix.

 

[rudransh verma]

It would not work if not correct. However I have modified to use the correct variables: $$(x-h)^2=4a(y-(d+a))$$ https://www.desmos.com/calculator/gq3qsmwreh

A glance at the parabola should tell you that 'k' the vertical shift parameter in your equation equals d+a where y=d is the equation of the directrix.

Nice!

 

[neilparker62]

$a$ is a variable, yes. One might call it a "parameter" that describes the parabola. It is the distance between the "vertex" of the parabola and its "focus".

A parabola can be described based on a given line (the "directrix") and a given point (the "focus"). The point is called the "focus". The parabola is the set of points that are at the same distance from the directrix as they are from the focus.

Thanks for this - had a hazy memory of focal point and directrix and I did not know that variable "a" in this form of the equation is the distance you mention above.

Perhaps worth noting that the "a" here corresponds to $\frac{1}{4a}$ in other standard forms such as $$y=ax^2+bx+c$$ or $$y=a(x-p)^2+q$$