# Algebra 2/Parabolas
## In Terms of Distances

Although we define the parabola as an intersection of a cone and plane parallel to the edge of the cone, this is pretty clearly a difficult definition to work with.
But luckily it turns out that there is a proof (due to Dandeline, which can be found in a nice presentation here: Dandeline's proof) that a parabola is equivalently defined as the locus of points equally distant between some point and some line.
More precisely, we begin with a point which we call the focus, and a line which we call the directrix. In principle these may be any point and line which are not coincident.
For simplicity we will assume (until indicated otherwise) that the focus has horizontal coordinate equal to 0. This will make calculations simpler. The only thing left unknown about the focus, then, is its vertical coordinate.
We will write the focus as $F(0,c)$.
Also for simplicity we assume that the directrix is horizontal. Therefore the directrix always has the form
    $y=a$ where $a\neq c$
We also define the vertex to be the point which is equally distant from the focus to the line. This makes the vertex either the minimum point (if the directrix is below the focus) or the maximum point (if the directrix is above).
* Solution
* A parabola is the locus of all points equally distant between a point and line.
* The diagram to the right shows essentially this situation, although coordinates are not shown explicitly.
* Still, if the focus is $F(0,2)$ and the vertex is visibly straight down from this by a distance of 2, then the vertex is $V(0,0)$.
* Solution
* Focus at (0,0), directrix at x=1.
* As with most things in algebra and geometry, your first instinct should always be to draw a picture to help understand the problem. I've drawn the picture to the right.
* As always with parabolas, the vertex is just precisely half-way between the focus and directrix.
* In this case, it is V(1/2, 0).
In general, a point $P(x,y)$ is on the parabola if and only if the distance FP equals the distance from P to the directrix.
Because F(0,c) then the distance FP is given by the distance formula,
    $d_{f}={\sqrt {(x-0)^{2}+(y-c)^{2}}}$
Because the directrix is horizontal, it is easy to find the distance to P.
    $d_{d}=|y-a|$
By definition of a parabola, being points at which these two distances are equal, then
    $d_{f}=d_{d}$
We define p to be the distance from the vertex to the focus and therefore, trivially, it is always half the distance from the focus to the directrix. It can always be computed by
    $p=|c-a|/2$
If we write the coordinates of the vertex as $V(h,k)$ then clearly $h=0$ because the focus, too, has its horizontal coordinate at 0. Also, if we assume $c>a$ then
    $k=c-p=a+p$
If on the other hand $c<a$ then $k=c+p=a-p$.
* Solution
* $d_{f}^{2}=x^{2}+(y-c)^{2}=x^{2}+y^{2}-2cy+c^{2}$
* and
* $d_{d}^{2}=(y-a)^{2}=y^{2}-2ay+a^{2}$
* so setting these equal, we can cancel the $y^{2}$ on each side,
* $x^{2}-2cy+c^{2}=a^{2}-2ay$
* Solving for $x^{2}$ we get
* $x^{2}=2y(c-a)+a^{2}-c^{2}$
* which, by the difference of squares, is
* $x^{2}=2y(c-a)+(a+c)(a-c)$
* Recalling that $c-a=2p$ and $a+c=2k$ then we get
* $x^{2}=4py+(2p)(-2k)$
* The right-hand side is $4py-4kp$. If we factor the $4p$ then we're done.
* $x^{2}=4p(y-k)$
* Solution
* Since $F(0,2)$ then we are given c=2. Since the directrix is $y=-2$ then we are given $a=-2$. Since there is no horizontal translation $h=0$.
* $k={\frac {a+c}{2}}={\frac {-2+2}{2}}=0$
* $p={\frac {c-a}{2}}={\frac {2-(-2)}{2}}=2$
* With these values known we can state the locations of all points.
* $F(0,2),V(0,0)$
* The directrix was given as $y=-2$.
* Finally, the equation of the parabola is $4\cdot 2(y-0)=x^{2}$ which simplifies to
* $8y=x^{2}$
* Solution
* We find many of the relevant quantities the same way as before. We have $c=10$ and $a=4$ so $k={\frac {a+c}{2}}={\frac {4+10}{2}}=7$. Also $p={\frac {c-a}{2}}={\frac {10-4}{2}}=3$. The equation would usually be $4\cdot 3(y-7)=x^{2}$ which is
* $12(y-7)=x^{2}$
* This problem is no different except that, because the focus is translated 2 to the left, then the relation is translated 2 to the left.
* We've seen earlier that to translate the relation $R(x,y)$ by $\langle i,j\rangle$, it is represented by $R(x-i,y-j)$. Here we translate by $\langle -2,0\rangle$.
* Therefore the equation of this translation is $12(y-7)=(x-(-2))^{2}$, which is
* $12(y-7)=(x+2)^{2}$

### The General Equation: Opens Upward

The above is a demonstration that the equation for a parabola which opens upward is
    $4p(y-k)=(x-h)^{2}$
As before, the equations $k={\frac {a+c}{2}}$ and $p={\frac {c-a}{2}}$.
a is the position of the directrix, $F(h,c)$ the coordinates of the focus, $V(h,k)$ the coordinates of the vertex, and p the distance from vertex to focus.
* Solution
* Since the general formula is $4p(y-k)=(x-h)^{2}$ then we can read off $h=1,k=-1$. We can see $4p={\frac {1}{4}}$ so that $p={\frac {1}{16}}$.
* Then $k={\frac {a+c}{2}}$ gives $-1={\frac {a+c}{2}}$. Also $p={\frac {c-a}{2}}$ gives ${\frac {1}{16}}={\frac {c-a}{2}}$. This produces the system of equations
* ${\begin{aligned}-2&=c+a\\{\frac {1}{8}}&=c-a\end{aligned}}$
* Summing the equations gives $-{\frac {15}{8}}=2c$ and therefore $c=-{\frac {15}{16}}$. Therefore
* ${\frac {1}{8}}=\left(-{\frac {15}{16}}\right)-a$
* which entails $a={\frac {17}{16}}$.
* Now every parameter has been determined.

### The General Equations: Downward, Rightward, Leftward

Until now we've assumed the directrix is horizontal, and the focus is above the directrix. This causes the parabola to open upward.
If the focus is below the directrix, all the same analysis applies. You will find that the only thing which changes is
    $p=-{\frac {c-a}{2}}$
which is necessary so that p is positive. This is necessary because p is a distance and so must be positive.
Then downstream of that, we get an equation which is only slightly different.
    $-4p(y-k)=(x-h)^{2}$
* Solution
* It's given that $h=1$ and $k=2$, and $c=0$. Then $k={\frac {a+c}{2}}$ gives
* $2={\frac {a+0}{2}}$
* so $a=4$. Also $p=-{\frac {c-a}{2}}$ gives
* $p=-{\frac {0-4}{2}}=2$
* Then the equation of the parabola is $4\cdot 2(y-2)=(x-1)^{2}$ which is
* $8(y-2)=(x-1)^{2}$
Next consider if the directrix is vertical rather than horizontal. Now the vertex and focus will share the same vertical coordinate by may have different horizontal coordinates. We now say that $F(b,k)$ and $V(h,k)$ are the coordinates of the focus and vertex, and as before p is the distance from focus to vertex.
Much like before,
    $h={\frac {a+b}{2}}$
If the focus is to the right of the directrix, the equations become
    $p={\frac {b-a}{2}}$
and the equation becomes
    $4p(x-h)=(y-k)^{2}$
If the focus is left of the directrix,
    $p=-{\frac {b-a}{2}}$
and the equation becomes
    $-4p(x-h)=(y-k)^{2}$
A consolidated display of the four cases is shown below.
vertical, horizontal / focus greater, lesser :
| Opens | Focus greater | Focus lesser |
| vertical | ${\begin{aligned}{\color {red}k={\frac {a+c}{2}}},&\quad \color {blue}p={\frac {c-a}{2}}\\4p(y-k)&=(x-h)^{2}\end{aligned}}$ | ${\begin{aligned}{\color {red}k={\frac {a+c}{2}}},&\quad \color {blue}p=-{\frac {c-a}{2}}\\-4p(y-k)&=(x-h)^{2}\end{aligned}}$ |
| horizontal | ${\begin{aligned}{\color {red}h={\frac {a+b}{2}}},&\quad \color {blue}p={\frac {b-a}{2}}\\4p(x-h)&=(y-k)^{2}\end{aligned}}$ | ${\begin{aligned}{\color {red}k={\frac {a+b}{2}}},&\quad \color {blue}p=-{\frac {b-a}{2}}\\-4p(x-h)&=(y-k)^{2}\end{aligned}}$ |
* Solution
* In effect we need to find the distance from the the vertex to the focus. The focus is the point where the rays intersect, and the vertex is the "bottom" of the depth of the parabola.
* It shouldn't matter where we place the satellite in our coordinate space, so just for simplicity, let's set the vertex at (0,0) and have the parabola open upward. So $h=0,k=0$. Then the equation is
* $4py=x^{2}$
* Because of this choice of coordinate system, the right edge of the rim is 10 feet to the right on the x axis, and 10 feet above the axis. That is to say, it passes through the point $(10,10)$.
* If we use these values in the equation,
* $4p\cdot 10=10^{2}$
* which implies $p=5/2$.
* Therefore the focus is 2.5 feet above the vertex.