More Analytic Geometry - The Hyperbola

Still not covered in our survey of plane analytic geometry is the conic section known as the hyperbola. Recall in the film 'Hidden Figures'  the most prolific human computer (Katherine Johnson) often had to compute when an elliptical orbit might become 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


Recall I also appended a note in the solutions to ellipse problems (Jan. 22):

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   =   Ö (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 shown 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 Problem:

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 th 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).      =   Ö5

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:

1) a) Factor the basic hyperbola equation and use this information to show how the asymptotes might be obtained. (Feel free to use any calculus concepts, including limits).

b) Label (on a graph) and identify the coordinates of the foci for the hyperbola:

y ²  / 9 ²   -    x ²  / 4 ²  = 1


2) In Problem #5 from the ellipse set, we saw an ellipse with equation:

x ²  /7  +    y ²   / 16 = 1

Change this to the form for a hyperbola, and:

a) sketch the graph

b) show the asymptotes

c) identify the foci coordinates

d) If the eccentricity of a hyperbola is expressed

e  =   Ö (a ²  +   b ²) /  a

Then label the directrices of the hyperbola in the graph if they are at distances (a/e) from the intersection of the asymptotes.

3) Analyze the equation below using the translation of axes approach:

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

Then:

a) Find the equation of the hyperbola with center at x' = y' = 0

b) The intercepts on the y' axis

c) the coordinates of the foci F1 and F2

d) The equations for the directrices

Seen 'Hidden Figures'? Here's A Look At Analytic Geometry (1)

In the new movie 'Hidden Figures', about three African-American women working at NASA who are "human computers" (computing spacecraft trajectories) there occurs a compelling moment  when Kevin Costner's character ("Al Harrison"- head of NASA's Langley computing section) asks if there isn't anyone who can do analytic geometry. At that point the white female supervisor (played by Kristen Dunst) points to Katharine G.  Johnson - who also chirps up "I can!"  This familiarity serves her well in computing orbits of satellites, manned craft as well as changing orbits - say from circular or elliptical to parabolic.

Given the currency of the film, it's relevant  to look at some examples from analytic geometry and offer up some problems too, What is analytic geometry? It is basically that area of mathematics that applies algebra and calculus to geometry in order to work out the locus of points for various figures, always expressible in the form F (x, y) = 0.

Once we generate the curve F(x,y) = 0 we can then study its peculiar geometric properties (for example those for a parabola differ from those for an ellipse) as well as find the equation of a curve once its geometric properties are known. All of these things we will do and I will challenge readers to do in assorted problems!

Probably the most straightforward equation and curve to begin examining is the circle. Let us consider one defined by the equation:

  x ²   +    y ²  =  4

This is very simple to plot as a graph and I invite readers to try. You will end up with the figure shown below:



This is actually an equation of basic form:  x ²    +    y²  =      r² 

where r denotes the radius. Since the square root of  4  (=  r²  ) is 2, then we see from the graph the radius is 2.

But this simple example is deceiving because it may give the impression that all equations of the circle consist of three terms, and all circles are centered at the origin (0, 0). In fact, most interesting cases are circles which are off -center, and also of more complicated equations.

Consider, for example, the equation:

x ²    +    y²   + 2x   - 4y  =  11

Is this a circle? Yes, it is, but we must work to get it into the correct form to analyze, or graph.

First, write the left side as:

x ²    +  2x  +  y²   - 4y 

Now complete the square for the expression in x and y to get:

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

Re-configure the entire equation paying attention to terms added to both sides:

(x ²    +  2x  + 1)   +    (y²   - 4y    + 4)   =  11 +   4   +   1 =   16

And if we now write in the most general form using coordinates (h, k) for the center:

(x - h)²    + (y - k ) ²   =  r ² 

We see at a glance, that h = -1 and k =2 with the radius r = the square root of 16 or 4.  We can check this by doing the graph:

Inspection of the above graph shows that indeed (h, k) = (-1, 2), the coordinates of the center and the radius is 4 units.

There is also another, longer way to approach the problem using a generic analytical expression, e.g.

Ax ²    +  Ay²   + Dx  + Ey  + F   = 0   (A not equal 0)

 This can be reduced to the earlier equation (x - h)²    + (y - k ) ²   =  r ²    

by completing the square, which is left as an exercise for the energetic reader.

When you do this you should obtain the general, analytical expression for the radius:

  r ²     = (D ²   +    E ²      - 4AF) /  4 A ²    

Let's check to see that this works out for the example problem we set, i.e.

x ²    +    y²   + 2x   - 4y  =  11

Rewritten as:

x ²    +    y²   + 2x   - 4y  -  11  = 0

So:  A = 1,  D = 2,   E = -4  and F = -11

Then:

r ²     = [2 ²   +    (-4) ²    - 4 (1)(-11))]  /  4  ²  

r ²     =  [4 + 16  + 44]/   4   =   64/  4   =   16

And since:  r ²     =  16 then r = 4  which  checks out.

The coordinate of the center will also be found to be:

h =  (-D/ 2A)   and k =  (-E/ 2A)

Again, checking this for the example:

h =   -2/ 2(1) =   -1   and k  =  -(-4)/ 2 =  4/2 = 2

so (h, k ) =   (-1, 2)

which again checks.

Problems for the math-minded (or readers looking for something to challenge them!)

1)  Complete the square for the general analytic equation:

Ax ²    +  Ay²   + Dx  + Ey  + F   = 0

To show how the radius r is derived, as well as the coordinates for the center of a circle not necessarily at (0,0)

2) Check the equation:  x ²    +    y²  =  4

to see that the general form also applies here.

3) Find the coordinates of the center of each of the following circles and the radius r, also sketch the circle:

a)  x ²    +    y²  - 2 y   =  3

b) 2x ²    +  2  y²   + x  + y = 0

c) x ²    +    y²  + 2x  = 8

4) Obtain the equation for the circle shown below: