Focal Property of a Hyperbola - Maple Programming Help

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Focal Property of a Hyperbola

Main Concept

A hyperbola consists of two open, disconnected curves called branches, which are mirror images of each other and resemble infinite bows. The points on these branches which are closest together, and thus closest to the center, are called vertices. The line segment between the two vertices is known as the transverse or major axis. Beyond the vertices on the same line as the major axis (lying further from the center) there are two points, E and F, known as the foci.

 

A hyperbola can be described as the locus of points for which the absolute value of the difference between the distances from any point P to each focus is a constant.

 

In particular, the difference between these distances is always equal to the length of the major axis.

The general equation for an "East-West opening hyperbola" is: xh2a2  yk2b2 = 1, while the general equation for a "North-South opening hyperbola" is: yk2a2  xh2b2= 1, where h, k is the center, a is the length of the semi-major axis (the distance from each vertex to the center), and b is the length of the semi-minor axis (the perpendicular distance from each vertex to each asymptote).

Derivation of the general equation from the focal property

For simplicity, let's say that the hyperbola is centered at 0, 0 with the following foci: E at c, 0 and F at c, 0. So, the distance from each focus to the center is c.

The distance from a general point Px, y to E is given by PE = x  c2+y  02 = x+c2+y2.

The distance from P to F is given by PF = x  c2+y 02= x  c2+y2.

Looking at the case in which P is a vertex of the hyperbola and subtracting the distances from this vertex to each focus, we see that the difference of these distances is 2a.

So, we know that:         

PEPF=2 a

x+c2+y2xc2+y2=2a

x+c2+y2 = 2a + xc2+y2

x+c2+y2 = 2 a + xc2+y22

x2+2cx+c2+y2=4a2+4 axc2+y2+xc2+y2

x2+2cx+c2+y2 = 4a2+4 axc2+y2 + x2  2 c x + c2 + y2

2c x = 4 a2+4 axc2+y2  2 c x

4c x  4 a2 =  4 a xc2+y2

c x  a2 =  a xc2+y2

c x  a22 = a2xc2+y2

c2x22a2cx+ a4 = a2x22 a2cx+ a2c2 +a2y2

c2x2+ a4 = a2x2+ a2c2 +a2y2

 a4a2c2=a2x2c2x2 + a2y2 

a2a2c2=a2c2x2+a2y2

Now, since the foci lie further from the center than the vertices, c > a, and so c2 > a2. We multiply by 1 to make both sides positive:

a2c2a2=c2a2x2a2y2 

Note that  c2 = a2 + b2, so b2 = c2  a2. Substituting b2, we get:

a2b2  = b2x2 a2y2

1=x2a2y2b2

This is the standard equation for a hyperbola centered at 0, 0 with semi-major axis length a and semi-minor axis length b. 

Asymptotes

Further from the center of the hyperbola, the branches approach, but never cross, two lines known as asymptotes. For a hyperbola with horizontal major axis, the general equation for these asymptotes is: y = ±baxh + k,

whereas if the major axis is vertical, the general equation for these asymptotes is: y = ±abx  h + k.

To prove that these equations represent the asymptotes of a hyperbola, we can look at the most basic case - a hyperbola with horizontal major axis centered at 0, 0 with semi-major axis length a and semi-minor axis length b - and can then extend this result to hyperbolae centered away from the origin.

We must prove that the line y = ba x is an asymptote of the hyperbola x2a2y2b2=1 in the first quadrant (all other quadrants follow by symmetry):

x2a2y2b2=1

x2a21=y2b2

b2 x2a21 = y2

b2x2a21=y

So, y = b2x2a21 is the equation of the hyperbola in the first quadrant.

Now, if y = ba x is an asymptote of the hyperbola as we claim it to be, this should be the line that the hyperbola approaches as x grows infinitely large.

We can take the limit of the difference between these functions as x  , and if this limit equals 0, then we can be certain that this line is an asymptote.

L = limxb2x2a21  bax

L = limxbax2  a2  x

L = limxbax2  a2  x2x2  a2 + x

L = limxba a2 x2  a2 + x

L = ablimx1x2  a2 + x

L = ab0

L = 0

Therefore the branch of the hyperbola, y = b2x2a21 approaches y = ba x as x   in the first quadrant.

 

Click to add two points to the plot below to set your foci, E and F. Then, choose the "Plot Points" radio button and click on the graph to plot the points creating a hyperbola. Select the "Show Hyperbola" check box to see the actual curve and its equation, and select the "Show Asymptotes" check box to view the asymptotes and their equations. Click "Reset" to reset the graph.

 

  

Distance from Focus E to Latest Point =

Distance from Focus F to Latest Point =

Difference between Distances = 

Equation of Hyperbola:

Equations of Asymptotes: