Monday, January 16, 2017

A Look At Analytic Geometry (3): Parabolas

From parabolic telescope mirrors to receivers, the parabola plays a major role, and so it is useful to study the properties of this particular figure. Probably the classic parabola of form:

y  =   x2   

is a good place to start. A graph of this equation is shown below:

No photo description available.

Note the vertex is on the origin or (0,0), but as with the circle and ellipse this need not be so, and one can encounter parabolas that are displaced, e.g. away from the x-axis, as well as oriented in a different way. For example, consider:  x2     = -  y  or    y  =  -   x2    

And the graph shown below:
No photo description available.

One can also obtain parabolas with vertices oriented to the left or right of the y-axis. It is left to readers to try to find equations that yield these.

The general form of the parabola is often given as:   x2  =  -  4py 

If the parabola opens downward as shown above. If the reverse is true it is given as: x2  =  4py  . In each case, the focus of the parabola is on the axis of symmetry (e.g. follow the y-axis through the vertex downwards - or upwards in the earlier case) and p units from the vertex.

Let's consider the parabola shown  for   x2  =  -  y    then:

- 4p =   1  and p = - 1/4

That is, -  1/4 unit from the vertex to get to the focus F. What would we have for the previous parabola? Well, you should be able to work out: p = 1/4

Now, the directrix of the parabola will be the line y = p or y =  -1/4 for the downward oriented parabola. It is a line parallel to the x-axis and by that amount of displacement.  It should therefore be no surprise that the tangent to a parabola at its vertex is parallel to the directrix.  Consider a parabola of form:

x2  =   4py 

Then the slope of the tangent at any point is dy/dx = x/ 2p. which is 0 at the origin.  But the second derivative:  d2 y /  dx2   = 1/2p which is positive.(So the curve is concave upward).

Of course, not all forms for the parabola are as straightforward as shown. Consider the equation:

2 x2  +   5y   - 3x + 4 = 0

We want to try to get this into the more general form for the parabola:

(x -   h) 2  =   - 4p (y -  k)

We can begin by dividing both sides of the initial equation by 2:

 x2  +   5y /2   - 3x/2  + 2   = 0

Which can also be written as:

x2     - 3x/2  =    - 5y /2   -  2

We can then complete the square, i.e. adding (- 3/4)2   to both sides:

x2     - 3x/2   + 9/16  =    - 5y /2   -  2   + 9/16

Or, write as:

(x  - 3/4)2     =   -5y/ 2  - 23/16

We can factor out 5/2 from the right hand side, i.e.:

(x  - 3/4)2     =   -5/ 2  (y   + 23/ 40) 

Now, compare this to the general form we provided:

(x -   h) 2  =  - 4p (y -  k)

From this we see that:  h = 3/4,  k =  -23/40  and 4p = 5/2  so: p = 5/8

Therefore, the following hold true and can be confirmed by sketching the graph (see problem 1):

a) The vertex is situated at V(3/4, - 23/40)

b) The axis of symmetry as always is x = h or x = 3/4 in this case.

c) The focus is p units below the vertex at F(3/4, - 6/5)


1) Graph the parabola (x  - 3/4)2    =   -5/ 2  (y   + 23/ 40)   and defined by the properties given above.

2)The parabola shown below:

No photo description available.
Has the general form  (y -   k) 2  =  - 4p (x  -  h)

Use this information to: a) find the actual equation of the curve, b) the coordinates of the focus, c) the equation for the directrix.

3) Find the focus and the directrix for the parabola: x2  =   8y

And graph it, clearly labeling the axis of symmetry.

Take the 2nd derivative before graphing and establish whether the parabola is oriented concave downward or upward.

4) Do the same for: y = x2  +   4x

5) Given  V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix.

6)  Find the tangent to the parabola y  =   x2     using the fact that the equation of the line tangent to a curve y = f(x) is given by:

y - y1 = f'(x1) (x - x1)

where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1  =   (x1)2     .  Show the graph of the given parabola and the tangent line with the point P1 (x1, y1) identified.

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