Parabolas

 Let d be a line in the plane and F a fixed point not on d. A parabola is the collection of points in the plane that are equidistant from F and d. The point F is called the focus and the line d is called the directrix.
 The picture illustrates the definition. The point P is a typical point on the parabola so that its distance from the directrix, PQ, is equal to its distance from F, PF. The point marked V is special. It is on the perpendicular line from F to the directix. This line is called the axis of symmetry of the parabola or simply the axis of the parabola and the point V is called the vertex of the parabola. The vertex is the point on the parabola closest to the directrix.  Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

(x-0)2+(y-p)2 = (y+p)2+(x-x)2.
x2+(y-p)2 = (y+p)2.
If we expand all the terms and simplify, we obtain
x2 = 4py.

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

x2 = 4py.
The graph of the parabola would be the reflection, across the x axis of the parabola in the picture above. A way to describe this is if p > 0, the parabola "opens up" and if p < 0 the parabola "opens down".

Another situation in which it is easy to find the equation of a parabola is when we place the focus on the x axis, the vertex at the origin and the directrix a vertical line parallel to the y axis. In this case, the equation of the parabola comes out to be

y2 = 4px
where the directrix is the verical line x=-p and the focus is at (p,0). If p > 0, the parabola "opens to the right" and if p < 0 the parabola "opens to the left". The equations we have just established are known as the standard equations of a parabola. A standard equation always implies the vertex is at the origin and the focus is on one of the axes. We refer to such a parabola as a parabola in standard position.

Parabolas in standard position
In this demonstration we show how changing the value of p changes the shape of the parabola. We also show the focus and the directrix. Initially, we have put the focus on the y axis. You can select on which axis the focus should lie. Also you may select positive or negative values of p. Initially, the values of p are positive.

If we translate a parabola in standard position both horizontally and vertically we would like to know the equation of the resulting parabola. We know that horizontal translations correspond to replacing x with x-c and vertical translations correspond to replacing y with y-d we obtain the following:

 Standard Equation Shift vertex to Resulting Equation x2=4py (h,k) (x-h)2=4p(y-k) y2=4px (h,k) (y-k)2=4p(x-h)

If we take the equation (x-h)2=4p(y-k) and expand it we get x2-2hx+h2=4py-4pk or x2-2hx-4py+4pk+h2=0 which is an equation of the form x2+Ax+By+C=0, where A, B and C are constants. The question we have is if we are given such an equation can we recognize it as the equation of a parabola? The answer can be determined by reversing these steps and this is achieved by completing the square.

 Example Determine if x2+4x+8y+12=0 is the equation of a parabola. If so, find the coordinates of the vertex and the focus and the equation of the directrix. Since only the x terms are quadratic, we only need complete the square on these terms. We get x2+4x+8y+12=0 (x+2)2-4+8y+12=0 (x+2)2+8y+8=0 (x+2)2+8(y+1)=0 (x+2)2=-8(y+1) (x+2)2=4(-2)(y+1) We should recognize this as a parabola that opens up and has been shifted 2 units to the left and 1 unit down. Thus the vertex is at (-2,-1). The value of p is -2. We know that in standard position the focus would be at (0,-2) but now that we have translated the curve the focus is located at (0-2,-2-1) or (-2,-3). The equation of the directrix in standard position is y = -(-2). Now we need to translate this 1 unit down. Then the equation of the shifted directrix is y+1 = 2 or y=1. The picture illustrates the shift of the parabola from standard position to the new position. Similarly, if we are given an equation of the form y2+Ay+Bx+C=0, we complete the square on the y terms and rewrite in the form (y-k)2=4p(x-h). From this, we should be able to recognize the coordinates of the vertex and the focus as well as the equation of the directrix.

In this next exercise set you are given the equation of a parabola in expanded form. On paper, determine the coordinates of the vertex and the focus. When you have your answers, press "Graph" to see the graph of the parabola and the coordinates of the vertex and focus.

Finding the Vertex and Focus of a Parabola