In previous lessons on conic sections, we discussed both the circle and the ellipse, which each result from "slicing" a cone clear through from leftright. In this lesson, we will discuss the shape formed when we slice through only one side of the cone, creating a bowlshaped figure called a parabola.
Consider the "hourglass" figure we used in the definitions of the circle and ellipse, created by connecting two infinite cones at their tips. What limitation would there be on the angle of the slice we would take out of one of the cones, if we wanted to only get a parabola (not get an ellipse, and not hit the other cone in any way)?
Watch This
Embedded Video:
 Khan Academy: Parabola Focus and Directrix 1
Guidance
We’ve examined ellipses and circles, the two cases when a plane intersects only one side of the cone and creates a finite crosssection. Is it possible for a plane to intersect only one side of the cone, but create an infinite crosssection?
It turns out that this is possible if and only if the plane is parallel to one of the lines making up the cone. Or, in other words, the angle between the plane and the horizon, is equal to the angle formed by a side of the cone and the horizontal plane.
In the image above, if you till the plane a little bit to the left it will cut off a finite ellipse (possibly a very large one if you only tilt it a little.) Tilt the plane to the right and it will intersect both sides of the cone, making a twopart conic section called a hyperbola, which will be discussed in the next section.
When the plane is parallel to from the side of the cone, the infinite shape that results from the intersection of the plane and the cone is called a parabola. Like the ellipse, it has a number of interesting geometric properties.
The equation of a parabola is simpler than that of the ellipse. There are a couple of methods of deriving the equation of a parabola, in this lesson we explore the distance formula:
The first method arises directly from the focusdirectrix property discussed in the previous section. Suppose we have a line and a point not on that line in a plane, and we want to find the equation of the set of points in the plane that is equidistant to these two objects. Without losing any generality, we can orient the line horizontally and the point on the \begin{align*}y\end{align*}axis, with the origin halfway between them. Since the parabola is the set of points equidistant from the line and the point, The parabola passes through the origin, (0,0). The picture below shows this configuration. The point directly between the directrix and the focus (the origin in this case) is called the vertex of the parabola. Suppose the focus is located at \begin{align*}(0,b)\end{align*}. Then the directrix must be \begin{align*}y = b\end{align*}.
Thus, the parabola is the set of points \begin{align*}(x,y)\end{align*} equidistant from the line \begin{align*}y = b\end{align*} and the focus point \begin{align*}(0,b)\end{align*}. The distance to the line is the vertical segment from \begin{align*}(x,y)\end{align*} down to \begin{align*}(0,b)\end{align*}, which has length \begin{align*}y  (  b) = y + b\end{align*}. The distance from \begin{align*}(x,y)\end{align*} to the focus \begin{align*}(0, b)\end{align*} is \begin{align*}\text{distance} = \sqrt {(x  0)^2 + (y  b)^2}\end{align*} by the distance formula. So the equation of the parabola is the set of points where these two distances equal.
\begin{align*}y + b = \sqrt{(x  0)^2 + (y  b)^2}\end{align*}
Since distances are always positive, we can square both sides without losing any information, obtaining the following.
\begin{align*} y^2 + 2by + b^2 & = x^2 + y^2  2by + b^2 \\ 2by & = x^2  2by \\ 4by & = x^2 \\ y & = \frac{1}{4b}x^2\end{align*}
But \begin{align*}b\end{align*} was chosen arbitrarily and could have been any positive number, and for any positive number, \begin{align*}a,\end{align*} there always exists a number \begin{align*}b\end{align*} such that \begin{align*}a = \frac{1}{4b}\end{align*}, so we can rewrite this equation as:
\begin{align*}y = ax^2\end{align*}
where \begin{align*}a\end{align*} is any constant.
This is the general form of a parabola with a horizontal directrix, with a focus lying above it, and with a vertex at the origin. If \begin{align*}a\end{align*} is negative, the parabola is reflected about the \begin{align*}x\end{align*}axis, resulting in a parabola with a horizontal directrix, with a focus lying below it, and with a vertex at the origin. The equation can be shifted horizontally or vertically by moving the vertex, resulting in the general form of a parabola with a horizontal directrix and passing through a vertex of \begin{align*}(h,k)\end{align*}:
\begin{align*}y  k = a(x  h)^2\end{align*}
Switching \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, the equation for a parabola with a vertical directrix and with a vertex at \begin{align*}(h,k)\end{align*} is:
\begin{align*}x  h = a(y  k)^2\end{align*}
Example A
Sketch a graph of the parabola \begin{align*}y = 3x^2 + 12x + 17\end{align*}.
Solution
First, we need to factor out the coefficient of the \begin{align*}x^2\end{align*} term and complete the square:
\begin{align*}y & = 3(x^2 + 4x) + 17 \\ y & = 3(x^2 + 4x + 4) + 17  12 \\ y & = 3(x + 2)^2 + 5\end{align*}
Now we write it in the form \begin{align*}y  k = a(x  h)^2\end{align*}:
\begin{align*}y  5 = 3(x + 2)^2\end{align*}
So the vertex is at \begin{align*}(2, 5)\end{align*} and plotting a few \begin{align*}x\end{align*}values on either side of \begin{align*}x = 2\end{align*}, we can draw the following sketch of the parabola:
Example B
Sketch a graph of the following parabola: \begin{align*}y = 2x^2  2x  3\end{align*}
Solution
Factor out the 2: \begin{align*}y = 2(x^2  x ) 3\end{align*}
Complete the square: \begin{align*}y + \frac{1}{2} = 2(x^2 x + \frac{1}{4}) 3\end{align*}
Add 3 to both sides and factor the trinomial: \begin{align*} y + 3 \frac{1}{2} = 2(x  \frac{1}{2})^2\end{align*}
The vertex (h, k) is: \begin{align*}(\frac{1}{2}, 3 \frac{1}{2})\end{align*}
Plot a couple of points to get:
Example C
Sketch a graph of the following parabola: \begin{align*}3x^2 + 12x + 11  y = 0\end{align*}
Solution
Factor out the 3 and move y and 11: \begin{align*}3(x^2 + 4x ) = y  11\end{align*}
Complete the square: \begin{align*}3(x^2 + 4x +4) = y 11 + 12\end{align*}
Factor the trinomial and collect like terms: \begin{align*}3(x + 2)^2 = y + 1\end{align*}
The vertex (h, k) is at: \begin{align*}(2, 1)\end{align*}
Plot a couple of points to get:
Concept question wrapup In order to only get a parabola cross section, the slice must be taken at the same angle as the side of the cone, that way the edge of the slice runs parallel to the edge of the cone and never intersects it at either top or bottom. This can be seen by a close look at the image from above:


Vocabulary
The focus point of a parabola is similar to the focus of an ellipse, representing one of the two locations whose sum total distance to any point on the curve is congruent.
Congruent means "exactly the same".
The directrix is a line that functions similarly to the second focus of an ellipse, as all points on the ellipse are the same sum total distance from the directrix and the focus.
Guided Practice
Questions
1) Sketch a graph of the following parabola: \begin{align*}0 = x^2  y + 3x + 5\end{align*}
2) Identify which of the following equations are parabolas:
 a) \begin{align*}y  5x + x^2 = 3\end{align*}
 b) \begin{align*}x^2  3y^2 + 3y  2x + 15 = 0\end{align*}
 c) \begin{align*}x  6y^2 + 20x  100 = 0\end{align*}
3) Calculate the distance between (3, 4) and (9, 5)
4) Calculate the distance between (2, 7) and (11, 23)
Solutions
1) Move the y, and complete the square, and factor to get: \begin{align*}(x + \frac{3}{2})^2 = y  2 \frac{3}{4}\end{align*} The focus is: \begin{align*}(\frac{3}{2}, 2 \frac{3}{4})\end{align*}
 Plot points to get:
2) Recall that a parabola has a squared input term only:
 Since: \begin{align*}x^2  3y^2 + 3y  2x + 15 = 0\end{align*} is squared on the input and output. In other words, both x and y are squared, therefore it is not a parabola.
 \begin{align*}y  5x + x^2 = 3 \end{align*} and \begin{align*}x  6y^2 + 20x  100 = 0\end{align*} are the correct answers.
 By the way, after completing the square, the first equation factors as: \begin{align*}(x  1)^2  3(y  \frac{1}{2})^2 + 14 \frac{3}{4} = 0\end{align*}
 Look familiar? It should, you worked on a bunch of them last chapter!
3) To calculate the distance between (3, 4) and (9, 5), use the distance formula \begin{align*}\text{distance} = \sqrt {(x_2  x_1)^2 + (y_2  y_1)^2}\end{align*}
 \begin{align*}\text{distance} = \sqrt {(9  3)^2 + (5  4)^2}\end{align*} ..... Substitute
 \begin{align*}\text{distance} = \sqrt {(6)^2 + (1)^2}\end{align*} ..... Simplify
 \begin{align*}\text{distance} = \sqrt {36 + 1} \to \sqrt{37}\end{align*}
\begin{align*}\therefore \sqrt{37}\end{align*} is the distance between (3, 4) and (9, 5)
4) To calculate the distance between (2, 7) and (11, 23), use the distance formula \begin{align*}\text{distance} = \sqrt {(x_2  x_1)^2 + (y_2  y_1)^2}\end{align*}
 \begin{align*}\text{distance} = \sqrt {(11  (2))^2 + (23  7)^2}\end{align*} ..... Substitute
 \begin{align*}\text{distance} = \sqrt {(13)^2 + (16)^2}\end{align*} ..... Simplify
 \begin{align*}\text{distance} = \sqrt {169 + 256} \to \sqrt{425} \to 5\sqrt{17}\end{align*}
\begin{align*}\therefore 5\sqrt{17}\end{align*} is the distance between (2, 7) and (11, 23)
Practice
Graph the following:
 \begin{align*}y = 3(x  1)^2 + 2\end{align*}
 \begin{align*}y = 3(x  1)^2  1\end{align*}
 \begin{align*}x = 2(y + 2)^2  1\end{align*}
 \begin{align*}y = 3(x + 4)^2  1\end{align*}
 \begin{align*}y = (x  3)^2\end{align*}
 \begin{align*}x = 3(y)^2 + 1\end{align*}
 \begin{align*}x = (y  3)^2  4\end{align*}
 \begin{align*}x = 3(y  4)^2 + 4\end{align*}
 \begin{align*}y = 2(x + 3)^2  3\end{align*}
 \begin{align*}x = (y  1)^2  2\end{align*}
For problems 1120, imagine a limited cone (not infinitely tall), as pictured below. Assume the two coordinates listed represent the intersection of a parabolic curve and the top of the cone. If the top surface of the cone were represented by the xaxis, then the two coordinates could be considered the xintercepts of the equation of the parabola. Find the distance between the points, and where required, the coordinates of the points.
 Coordinates: (20, 17) and (6, 1)
 Coordinates: (1, 5) and (6, 13)
 Coordinates: (1, 2) and (5, 5)
 Coordinates: (13, 12) and (15, 6)
 Coordinates: (3, 9) and (6, 14)
 \begin{align*}25x^2 + 15x + 10 = 0\end{align*}
 \begin{align*}24x^2 + 22x 4 = 0\end{align*}
 \begin{align*}4x^2 24x + 32 = 0\end{align*}
 \begin{align*}24x^2 + 54x + 27 = 0\end{align*}
 \begin{align*}12x^2 + 25x + 12 = 0\end{align*}