Analytic Geometry Revisited (2)

Parabola:

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:

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    

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 

Provided 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 above  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 begin by dividing both sides of the first equation by 2:

 x²  +   5y /2   - 3x/2  + 2   = 0

Which can 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:

(x -   h) ²  =  - 4p (y -  k)

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

Therefore, the following hold true and can be confirmed by sketching the graph. :

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)

Hyperbola:

In the 2016 film 'Hidden Figures'  the most prolific human computer (Katherine Johnson) often had to compute when an elliptical orbit  became hyperbolic. In the case of the ellipse we saw that the equation for this curve in Cartesian coordinates was:

x ²  / a ²  +    y ²  / b ² = 1

Note: In working these problems it also helps to realize that the ellipse can't have an e -value greater than 1 (which would make it a hyberbola) and one also can't  have imaginary values for the radical defining c.

Recall the equation relating semi-major (a) and semi-minor (b) axis to identify location of the foci was:

c   =    (a ²  -   b ²)

But in the case of the hyperbola this will become"

b  =   Ö (c ²  -   a ²)

Given now c is greater than a.

The equation for the hyperbola in Cartesian coordinates is:

x ²  / a ²   -    y ²   / b ²  = 1

A sketch of a representative hyperbola:

(x ²  / 9 ²   -    y ²   / 4 ²  = 1)

Is given below:

As seen here the hyperbola, like the ellipse, is symmetric with respect to both axes and the origin but it has no real y-intercepts. In fact, no portion of the curve lies between the lines x = a and x = -a, or in this case x = 3, and x = -3.  The equation of the asymptotes is also easily obtained, i.e. y = -bx/ a and y = bx/a, or y = -2x/ 3 and y = 2x/ 3.

As we see from this example, the two foci (F1 and F2) lie on the x-axis. Obtaining the coordinates of the foci is straightforward and is just:  F2 (+c, 0) and F1(-c, 0) where:

c =  Ö (a ²  +    b ²). In this case, c =   Ö (9  +    4).  

=    Ö 13.    F1 is then at (+ Ö 13, 0) and F2 is at

(- Ö 13 ,  0)  where Ö 13  =  3.6. 

We can also interchange x and y in the basic hyperbola equation, changing it to a hyperbola with its foci on the y-axis instead of the x -axis. In other words, we now write:

y ²  / 9 ²   -    x ²  / 4 ²  = 1

The graph for this hyperbola is shown below, with the equations identified for the straight (asymptote) lines:

Note the center of a hyperbola is the point of intersection of its axes of symmetry. For greater generality, if the center is at (h, k) we can also introduce a translation to new coordinates such that:

x' = x - h and y' = y - k

with origin O' at the center. Then in terms of the new (translated) coordinates, the equation for the hyperbola will be one of the following:

1) x' ²  / a ²   -    y' ²  / b ²  = 1

2)  y' ²  / a ²   -    x' ²   / b ²   = 1

Example:

Use the translation of axes technique to analyze the equation:

x ²  -   4  y  ²   + 2 x   +  8 y -   7 = 0

And thence find the equation for a hyperbola with center at x' = 0, y' = 0 and also identify the values of (x, y). Obtain the coordinates of the foci F1 and F2 and sketch the curve, label the foci and show the asymptotes.

Solution:

As we did before (ellipse, parabola) complete the squares in the x, y terms separately and reduce to standard form, so:

(x²   +   2 x)  +  4 (y²    - 2 y)  =  7

(x²   +   2 x  +  1)  +  4 (y²    - 2 y   + 1)  =  7 + 1  - 4  

Then:

(x + 2) ²  / 4 +    (y - 1 ) ²      = 1

Applying translation of axes:

x' = x + 1, and y' = y  - 1

Thus reducing the equation to:

x' ²  / 4  -    y' ²   / 1  = 1

Which represents a hyperbola with center at x' = 0 and y' = 0, or x = -1 and y = 1.

Then:   c =  Ö  (a ²  +    b ²).      =   Ö 5

The sketch of the hyperbola is shown below with foci and asymptotes identified:

It can easily be verified that the straight lines (asymptotes) have equations: y = x/ 2 and y = -x/2

Problems for the Math Maven:

1)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. 

2)  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.

3)Analyze the equation below using the translation of axes approach and graph the resulting curve:

x ²   -   4  y ²    -  2 x   + 8 y -  2 = 0