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I have trouble grasping parabolas, and mainly the cartesian equations describing the,. In my mind, there are 4 possible parabolas, a parabola shaped like a mountain ($\cap$), a parabola shaped like a valley ($\cup$), a parabola shaped like the greater than sign ($\supset$) and a parabola shaped like the smaller than sign ($\subset$), of course without the sharp edges like they have in my examples. I know one standard equation:

$y-b = \dfrac{1}{4c}(x-a)^2$, with the vertex at $T(a,b)$ and the focus $F(a; b+c)$.

However, I don't know for which of the 4 this one is, so my questions are:

1. What is the equation of the $\cap$ shaped parabola?

2. What is the equation of the $\cup$ shaped parabola?

3. What is the equation of the $\subset$ shaped parabola?

4. What is the equation of the $\supset$ shaped parabola?

To clarify: I would like them in the same form as the one I noted above in the example, thanks in advance.

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  • $\begingroup$ If the parabola has a vertical directrix, it is of the form $(y-k)^2=x-h$, if it has an horizontal directrix, it is of the form $(x-k)^2=y-h$. The vertex in the first case is $(h,k)$ and is $(k,h)$ in the other. $\endgroup$
    – Pedro
    Nov 8, 2012 at 19:10
  • $\begingroup$ @PeterTamaroff Only if the opening is towards the positive side... $\endgroup$
    – ronno
    Nov 8, 2012 at 19:11
  • $\begingroup$ @ronno Then take $-x$ or $-y$. I forgot to add that. $\endgroup$
    – Pedro
    Nov 8, 2012 at 19:14

4 Answers 4

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This general equation you gave is a parabola that opens either up or down (U or ^ shaped, in your words) depending on the signs of $c$: if $c > 0$ it opens up, if $c < 0$ it opens down. You can get the formulas for the right/left ones (C and >) by switching $x$ and $y$, and again whether it opens right or left depending on whether $c$ is positive or negative, respectively. There many other sorts of parabolas too, though--for example, you might have a parabola at a 45-degree angle to the axes.

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  • $\begingroup$ So it would be: $ (x-a) = \dfrac{1}{4c} (y-b)^2 ?$, and the vertex still at $(a,b)$ but the focus at $(a+c, b)$? $\endgroup$ Nov 8, 2012 at 19:14
  • $\begingroup$ Yup--exactly right. $\endgroup$ Nov 8, 2012 at 19:20
  • $\begingroup$ How can $c$ be less than zero, a distance can never be less than zero.. $\endgroup$ Nov 8, 2012 at 20:23
  • $\begingroup$ @ZafarS Think about it... You have $4c(x-a) = (y-b)^2.$ As $c$ changes sign the parabola changes from being open to the left to being open to the right. When $c=0$ you have a degenerate case: a double line. $\endgroup$ Nov 8, 2012 at 21:26
  • $\begingroup$ @FlybyNight I'm sorry, I thought $c$ was a distance, but is it just a coordinate? $\endgroup$ Nov 8, 2012 at 21:29
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Parabolae can come in all shapes and sizes. Draw yourself the standard parabola $y=x^2$ on a piece of paper. Now rotate that piece of paper. They are all still parabolae, just at different angles.

For the four parabolae that you mention, you have $y=x^2$, $y=-x^2$, $x=y^2$ and $x=-y^2$. (These can be moved around and stretch, for example $y = 2(x-1)^2 + 4$ will look a bit like $y=x^2$.

Let's think of the bigger picture. Parabolae are example of conics. Conics include ellipses, parabloae, hyperbolae, and other degenerate examples like double lines and crossing lines. The equation of a general conic is:

$$ax^2 + 2hxy + 2gx + by^2 + 2fy + c = 0$$

where the $a,h,g,b,f,$ and $c$ are numbers of your choice. The curve $y = x^2$ has $a=-1,$ $h = 0$, $g = 0$, $b=0$, $f = \frac{1}{2}$ and $c=0$. You can choose any $a,h,g,b,f,$ and $c$ and, provided $h^2 - ab = 0$, you will have the equation of a parabola.

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A parabola is the locus of points equidistant from a point and a line.

What's the distance from a location $r$ to a point labeled $a$? $|r-a|$.

What's the distance from a location $r$ to a line parallel to a unit vector $\hat \ell$? $|r-(r\cdot \hat \ell) \hat \ell|$.

Set the two equal, and you're done.

Example: $a = d \hat y$ and $\hat\ell = \hat x$. See that $r \cdot \hat \ell = x$.

$$\begin{align*}|r-a|^2 &= |r-(r\cdot \hat \ell) \hat \ell|^2 \\ x^2 + (y-d)^2 &= x^2 + y^2 -2 x^2 + x^2 \\ x^2 &= 2dy - d^2 \\ x^2 &= 2d\left ( y - \frac{d}{2} \right) \end{align*}$$

It all ends up falling out from the basic idea of being equidistant from a line and a point.

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  • $\begingroup$ Thank you, that's clear. However, my concern is, if you want to write these in the way I wrote them, You can get the $\cap$ form when $c<0$. However, isn't $c$ the distance between the vertex and directrix, how can a distance be negative? Or should $c$ be perceived as a coordinate? $\endgroup$ Nov 8, 2012 at 21:52
  • $\begingroup$ Interpreting $c$ as a coordinate should work fine. $\endgroup$
    – Muphrid
    Nov 8, 2012 at 21:57
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When you differentiate the function and draw a variation table:

aka. f(x)=y

f(x)=x^2 - 1 f'(x)=2x

x ... 0 ... f'(x) - 0 + f(x) -1 Thus it would have a "smiley-face shape" You can do this for all the other equations as well, but basically, if y=mx^2 When m>0 It will have a smiley-face shape When m<0 It will have a sad-face shape If x=my^2 When m>0 )-shape When m<0 (-shape

Hope this helps

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