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

[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 $$y=a(x-p)^2+q$$