Parabolas were introduced in the
Algebra II curriculum;
study/review the earlier lessons listed below.
Be sure to check your understanding by doing some of each problem type in the exercises.
 Parabolas
You must know the:
 definition of a parabola as the set of points in a plane
that are the same distance from a fixed point and a fixed line
 terminology: focus (the fixed point); directrix (the fixed line); vertex
 reflecting property of parabolas
 collecting property of parabolas

Equations of Simple Parabolas
 Know how to derive the equation ‘$\,x^2 = 4py\,$’ using the definition.
The equation $\,x^2 = 4py\,$ is one of the two standard forms for a parabola.
The other standard form, $\,y^2 = 4px\,$, is derived on this page (below).
 The parabola described by $\,x^2 = 4py\,$ is a function of $\,x\,$;
it can be equivalently written as $\displaystyle\,y = \frac{1}{4p}x^2\,$.
 Know how to use graphical transformations to
describe all parabolas with directrix parallel to the $x$axis.
 Know the significance of the parameter $\,p\,$ in the equations for a parabola.
 Be able to graph the equation $\,x^2 = 4py\,$ (and shifted versions).
In particular, be able to:
 determine if the parabola is concave up or concave down
 find the coordinates of the focus
 find the equation of the directrix
 plot an additional point that gives a sense of the ‘width’ of the parabola
You can play with parabolas at right:
 Move the focus.
How does the shape change as the focus moves closer to the directrix?
How does the shape change as the focus moves farther away from the directrix?
 Move the directrix.
When the directrix is parallel to the $y$axis, which way does the parabola open?
When the directrix is parallel to the $x$axis, which way does the parabola open?


Below is some information that wasn't covered in the Algebra II curriculum.
The exercises on this page address only this new information.
However, it is assumed that you've mastered the exercises from the two earlier parabola lessons
(Parabolas and
Equations of Simple Parabolas).
Axis of Symmetry for a Parabola
Recall that a parabola is determined by two pieces of information:
 a line (the directrix)
 a point not on the line (the focus)
The
axis of symmetry for a parabola is the line through the focus
that is perpendicular to the directrix.
If the parabola is ‘folded’
along this line, then ‘half’ the parabola perfectly
coincides with the other half.
All conic sections have (at least one) axis of symmetry.
For the
standard forms of conics, the axes of symmetry are always
the $x$axis and/or the $y$axis.
For the standard form $\,x^2 = 4py\,$ of a parabola, the axis of symmetry is the $\,y\,$axis.
For the standard form $\,y^2 = 4px\,$ of a parabola (derived next), the axis of symmetry is the $\,x\,$axis.
Deriving the standard form $\,y^2 = 4px\,$ for a parabola
Place a parabola with its vertex at the origin, as shown at right.

If we put the focus on the $\,x$axis,
then the directrix will be parallel to the $\,y$axis.
Why?
The axis of symmetry passes through the vertex (the origin) and the focus, hence is the $x$axis.
The directrix is perpendicular to the axis of symmetry.

Or, if we put the directrix parallel to the $\,y$axis,
then the focus will be on the $\,x$axis.
Why?
The axis of symmetry is perpendicular to the directrix and contains the vertex (the origin),
hence is the $x$axis.
The focus always lies on the axis of symmetry.
In either case, let
$\,p \ne 0\,$ denote the $\,x$value of the focus.
Thus, the focus has coordinates $\,(p,0)\,$.
Although the sketch at right shows the situation where
$\,p\gt 0\,$,
the following derivation also holds for $\,p \lt 0\,$.


Since the vertex is a point on the parabola,
the
definition of parabola
dictates that it must be the same distance from the focus and the directrix.
Thus, the directrix must cross the $\,x\,$axis at $\,p\,$;
indeed, every $\,x$value on the directrix equals $\,p\,$.
Let $\,(x,y)\,$ denote a typical point on the parabola.
The distance from $\,(x,y)\,$ to the
focus $\,(p,0)\,$ is found using the
distance formula:
$$
\tag{1}
\sqrt{(xp)^2 + (y0)^2 }
= \sqrt{(xp)^2 + y^2}
$$
To find the distance from $\,(x,y)\,$ to the directrix, first drop a perpendicular from $\,(x,y)\,$ to the directrix.
This perpendicular intersects the directrix at $\,(p,y)\,$.
The distance from $\,(x,y)\,$ to
the directrix is therefore the distance from $\,(x,y)\,$ to $\,(p,y)\,$:
$$
\tag{2}
\sqrt{(x(p))^2 + (yy)^2}
=
\sqrt{(x+p)^2}
$$
From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal:
[beautiful math coming... please be patient]
$$
\sqrt{(xp)^2 + y^2} = \sqrt{(x+p)^2}
$$
This equation simplifies considerably, as follows:
Squaring both sides: 
$
(xp)^2 + y^2 = (x+p)^2
$

Multiplying out: 
$
x^2  2px + p^2 + y^2 = x^2 + 2px + p^2
$

Subtracting $\,x^2 + p^2\,$ from both sides:

$
y^2  2px = 2px
$

Adding $\,2px\,$ to both sides: 
$
y^2 = 4px
$

This is the second standard form for a parabola, where the axis of symmetry is the $x$axis.
The most critical thing to notice is the coefficient of $\,x\,$, since
it holds the key to locating the focus of the parabola.
As an example, consider the equation $\,y^2 = 10x\,$.
Comparing with $\,y^2 = 4px\,$,
we see that $\,10 = 4p\,$, or $\,p = \frac{10}4 = \frac 52\,$.
Thus, $\,y^2 = 10x\,$ graphs as a parabola with vertex at the origin
and focus $\displaystyle\,(\frac 52,0)\,$.
Standard Forms for Parabolas
In summary, we have:
STANDARD FORM OF PARABOLA 
AXIS OF SYMMETRY 
VERTEX 
FOCUS 
DIRECTRIX 
$x^2 = 4py$ 
$y$axis 
$(0,0)$ 
$(0,p)$ 
$y = p$ 
$y^2 = 4px$ 
$x$axis 
$(0,0)$ 
$(p,0)$ 
$x = p$ 
Sketching Parabolas
You do not need to memorize lots of things to sketch
parabolas in standard form!
The key thing to memorize is that the important coefficient is $\,4p\,$,
where the size (absolute value) of $\,p\,$ gives
the distance from the focus to the origin.
You'll be letting the equation tell you the proper shape,
as indicated next:
 recognize form:
Recognize that you have an equation of the form $\,x^2 = 4py\,$ or $\,y^2 = 4px\,$.
If needed, find the discriminant to jog your memory: $\,B^2  4AC = 0\,$ indicates a parabola.
There's only one interesting coefficient in these equations ($\,4p\,$), so it is referred to
as ‘the coefficient’ below.
 isolate squared term:
Isolate the squared term.
That is, get the squared term (with a coefficient of $\,1\,$) all by itself on one side of the equation.
 determine shape (opening up/down/left/right):
Since a perfect square is always nonnegative, the other side of the equation must also be nonnegative.
Depending on the equation, this will force one of the following four situations:
 all the $\,x$values positive (parabola opens to the right)
 all the $\,x$values negative (parabola opens to the left)
 all the $\,y$values positive (parabola is concave up)
 all the $\,y$values negative (parabola is concave down)
The examples below illustrate how this works.
Use this information to (lightly) sketch the proper shape.
Remember that the vertex is the origin.
 plot the focus:
The focus is always inside the parabola.
To find its distance from the origin, throw away any minus signs (as needed) on the coefficient and divide by $\,4\,$.
Why? The coefficient is $\,4p\,$, and $\,p\,$ gives the distance from the focus to the origin.
Or, you can set the coefficient to $\,4p\,$ and solve for $\,p\,$ (which will include the correct sign).
 draw in the directrix:
Sketch the directrix.
It is the same distance from the origin as the focus,
on the opposite side.
 plot easy points to get the proper ‘width’:
There are always three easy points to get on a parabola,
once you know the focus and the directrix.
Refer to the sketch at right:
 The vertex is always halfway between the focus and directrix.
In the standard forms, the vertex is the origin.
 Use two fingers to mark the distance from the focus to
the directrix (shown in yellow at right).
Keeping one finger on the focus, sweep around ($\,90^\circ\,$ in each direction) to get
the two ‘easy points’. Why are they points
on the parabola? Their distances to the focus and directrix
both equal the radius of the circle shown (see the red segments
at right).
These two easy points give a sense of the ‘width’ of the parabola.

three easy points on a parabola:
the vertex;
two points to give a sense of the ‘width’

Example
Graph $\,y^2 = 3x\,$.
Solution:
 The equation $\,y^2 = 3x\,$ is in the form $\,y^2 = 4px\,$.
 The $\,y^2\,$ term is already isolated, with a coefficient of $\,1\,$.

$y^2\ge 0 \implies 3x\ge 0 \implies x\ge 0$
Thus, the $x$value of every point on the parabola is nonnegative;
the parabola opens to the right.

$4p = 3 \implies p = \frac 34\implies \text{focus } = (\frac 34,0)
$
 equation of directrix: $\,x = \frac 34\,$

two easy points:
distance from focus to directrix: $\,2(\frac 34) = \frac 32\,$
coordinates of easy points: $\,(\frac 34,\frac 32)\,$ and $\,(\frac 34,\frac 32)\,$
It's confidenceboosting to check that these points satisfy the equation $\,y^2 = 3x\,$:
$$\begin{gather}
\bigl(\pm \frac 32\bigr)^2\ \ \overset{?}{=}\ \ 3\bigl(\frac 34\bigr)\cr\cr
\frac 94 = \frac 94\qquad \text{Yep!}
\end{gather}
$$


Example
Graph $\,2x^2 + 5y = 0\,$.
Solution:
 The equation $\,2x^2 + 5y = 0\,$
has only $\,x^2\,$ and $\,y\,$ terms.
It can be put in the form $\,x^2 = 4py\,$.
 Isolate the square term, and get a coefficient of $\,1\,$:
$$
\begin{gather}
2x^2 + 5y = 0\cr
2x^2 = 5y\cr
x^2 = \frac 52y
\end{gather}
$$

$x^2\ge 0 \implies \frac 52y\ge 0 \implies y\le 0$
Thus, the $y$value of every point on the parabola is negative (or zero);
the parabola opens down.

$4p = \frac 52 \implies p = \frac 58\implies \text{focus } = (0,\frac 58)
$
 equation of directrix: $\,y = \frac 58\,$

two easy points:
distance from focus to directrix: $\,2(\frac 58) = \frac 54\,$
coordinates of easy points: $\,(\frac 54,\frac 58)\,$ and $\,(\frac 54,\frac 58)\,$
It's confidenceboosting to check that these points satisfy the equation $\,2x^2 + 5y = 0\,$:
$$\begin{gather}
2\bigl(\pm \frac 54\bigr)^2 + 5(\frac 58)\ \ \overset{?}{=}\ \ 0\cr\cr
2\bigl(\frac {25}{16}\bigr)  \frac{25}{8}\ \ \ \overset{?}{=}\ \ 0\cr\cr
0 = 0\qquad \text{Yep!}
\end{gather}
$$

