Parabolas were introduced in the
Algebra II curriculum.
Links to these earlier lessons are given below,
together with concepts that you must know from each section.
Be sure to check your understanding by doing some of each problem type in these earlier exercises.
These concepts are then further explored in this current lesson.
You can play with parabolas at right:


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).
Recall that a parabola is determined by two pieces of information:
Place a parabola with its vertex at the origin, as shown at right.
Thus, the focus has coordinates $\,(p,0)\,.$
Although the sketch at right shows the situation where
$\,p\gt 0\,,$


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: $$ \cssId{s70}{\sqrt{(xp)^2 + (y0)^2 } = \sqrt{(xp)^2 + y^2}} \tag{1} $$
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 $\,(p,y)\,$:
$$
\tag{2}
\cssId{s75}{\sqrt{(x(p))^2 + (yy)^2}
=
\sqrt{(x+p)^2}}
$$
From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: $$ \cssId{s77}{\sqrt{(xp)^2 + y^2} = \sqrt{(x+p)^2}} $$
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 $\displaystyle\,(\frac 52,0)\,.$
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  $(0,0)$  $(p,0)$  $x = p$ 
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:
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; two points to give a sense of the ‘width’ 
Graph $\,y^2 = 3x\,.$
Solution:


Graph $\,2x^2 + 5y = 0\,.$
Solution:


On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 