Variation Analysis, Hyperbola


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Hyperbola: Variation Analysis   


     Here we investigate the relationships between the parameters of the conic sections and the sections themselves. When the conic is centered at the origin its general equation is called its central equation.


central equation:

     The analysis for the hyperbola is similar to the ellipse. The quantities of interest are again, a, b, and c. The hyperbola has a similar eccentricity as the ellipse, namely

                                          ε =

Suppose we keep a constant and increase b.

For a given x in this case, then, y must increase.

Note: as b↑ ↓ which means ↓


Since we're keeping a constant and x fixed we're subtracting a smaller


from which breaks the equality. To maintain the equality we must counter the decrease in by multiplying by a larger . This


means that y2 must increase.

(start with 9 - 8 = 1;    if 8 = 32 / 4   and  8 = 48 / 6,  that means if  we allow 4 to increase to 6

then 32 must also increase to 48    so that we maintain the equation 9 - 8  = 1)


   



     Say x equals 6. Then, as b increases, the corresponding y value for x = 6 jumps to the next hyperbola; from the purple hyperbola to the red hyperbola then to the green and finally to the cyan hyperbola. In other words, as b increases the branches of the hyperbola open up.

     The eccentricity is written next to each hyperbola. Notice how the eccentricity increases as the branches open. The foci are also shown. Notice that as the branch opens, its focus moves away from the vertex.



     Now let's consider the effect of keeping b constant and allowing a to increase.

What we have then, is:


As argued previously decreases. So to keep the difference equal to 1 we must have

decrease as well, therefore y must decrease.

(once a gain start with  9 - 8  = 1;  say 8 = 32 / 4.  NOw if we decrease 4 to 2, then 32 must decrease to 16 to maintain  9  -  8 = 1.)

     The following graph shows that for a given x as a increases then y jumps from one hyperbola to the other which does not open as much. Let x = 9, then as a increases, then y moves from the purple to the red then to the green hyperbola. (The vertex of the blue hyperbola is beyond x = 9, so it plays no part, there is no y for x = 6 on the blue hyperbola.) The eccentricities are shown. Notice that they increase as the branches open further (as before.) Also note that the foci move toward the vertex as the eccentricity decreases.

( ε = c/a, so as ε ↓, c ↓ )



 

     Now let's look at the previous graph in terms of the a, b, c right triangles that define the asymptotes of hyperbolas.

     For each hyperbola b is the same. As we increase a from the purple triangle to the blue triangle c also increases. The asymptotes lie on each hypotenuse c. Their slopes decrease as we move across these triangles. The branches of the hyperbola collapse.



 

  

     Recall that we use these triangles to draw the asymptotes for the hyperbolas. And remember we're keeping b constant. As we increased a, the bases of these triangles, the hypotenuse c also had to increase. The hypotenuse of any right triangle is always greater then either of its legs, so c > a and therefore the ratio c/a > 1. So this means that for hyperbolas

                                  ε > 1

     As a → c (and therefore ε 1) the branches of the hyperbola collapse to the X axis. Likewise, we can say (from the analysis with a constant and b changing) that as b → c (and therefore ε becoming very large, without bound) then the branches of the hyperbola collapse toward the Y axis.

     a2 + b2 = c2 → a2= c2 - b2 → when b gets very close to c, a2 becomes very small, very close to zero, so c/a becomes very very large.


     Next is a graph of the asymptotes of these hyperbolas. Remember that each pair of colored asymptotes define the boundaries of the corresponding hyperbola. As ε the branches open.



 


Eccentricity Analysis for the Hyperbola


     As with the ellipse we want to find a relationship between c and a for all hyperbolas. All hyperbolas have a directrix. It is a line defined at the distance a2/c from the center of the hyperbola and is perpendicular to the axis containing the vertices; There is directrix for each branch.

     The first diagram shows the directrix and the quantities of interest. The second diagram pulls the triangle out to make the magnitudes of the legs a bit clearer and we'll use this diagram for our analysis.





  
     



 


     As with the ellipse, our goal is to find a relationship between d, the distance from the point on the hyperbola to the directrix and r1, the distance from this point to the focus f1. This relationship we want to be in terms of c and a only.

    The common leg between both right triangles is labeled w. Using Pythagorean's theorem on the smaller triangle we get

w = (r1)2 - (c - x)2 = (r1)2 - c2 + 2cx - x2

On the larger triangle we get

w = (r2)2 - ( c + x )2 = (r2)2 - c2 - 2cx - x2

 

     Now setting the two equations equal we get

(r1)2 - c2 + 2cx - x2 = (r2)2 - c2 - 2cx - x2

Simplifying we get

(r1)2 - (r2)2 = 4cx

     Recall the defining property of a hyperbola in this case is r2 - r1 = 2a

First factor

(r1 - r2) (r1 + r2) = 4cx

     Now divide the left hand side by r2 - r1 and the right hand side by 2a


And we get

r1 + r2 = 2cx/a

Using r2 - r1 = 2a, once for r1 and again for r2

r1 + (r1 + 2a) = 2cx/a (r2 - 2a) + r2 = 2cx/a

2r1 = cx/a - 2a 2r2 = 2cx/a + 2a

r1 = 2cx/2a - 2a/2 r2 = 2cx/2a + 2a/2

r1 = cx/a - a r2 = cx/a + a

We're almost there.

d = x - a2/c

Now we want to relate d to r1

so solve for x in r1 = cx/a - a and substitute that into the previous equation for d.

r1 + a = cx/a

a r1 + a2 = cx

(ar1 + a2)/c = x ------> d = (ar1 + a2)/c - a2/c

d = (ar1)/c + a2/c - a2/c

d = ar1/c

finally divide by r1

d/r1 = a/c

The ratio we want is r1/d, so invert both sides and we get

r1/d = c/a = ε the eccentricity of the hyperbola.

 
 

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