Solutions to Hyperbola 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).

Solution:

We have:

x ²  / a ²  +    y ²  / b ² = 1

Factor to get:

(x/a + y/b) (x/a - y/b) = 1

Take limit for one of the factors, viz:

  lim x _®oo    (x/a - y/b)   =  0
        y _®oo   

Which leads one to conjecture that the straight line:

(x/a - y/ b) = 0  may be an asymptote for the curve.

Rewrite:   x/ a = y/ b   or  y =  bx/ a

And we see it comports with the example in the text, e.g. 

y = bx/a,  or (for specific case)  y = -2x/ 3 and y = 2x/ 3.

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

y ²  / 9    -    x ²  / 4 ²  = 1

Solution:

In this case the two foci (F1 and F2) lie on the y-axis. Obtaining the coordinates of the foci is straightforward and is just:  F2 (0, +c) and F1(0, -c) where c =  Ö 13   or 3.6. So the labeled graph is shown below:



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.

Solution:

We write the equation for the relevant hyperbola:

x ²  /7  -    y ²   / 16 = 1

The graph with the asymptotes and foci show is sketched below:

The foci coordinates are defined at  F1 (+c, 0) and F2 (-c, 0) where in this case, c =
Ö 23  =  4.8. 

The foci coordinates then are at: F1 ( Ö 23   , 0 ) and F2 (-Ö 23  , 0)

The eccentricity would be:  e  =   Ö (a ²  +   b ²) /  a

Or:   e  =  Ö 23  / Ö7  =    1.81 

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

Solution:

We rewrite the equation:

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

Then:

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

At first one might be tempted to divide through by 4, but this isn't useful because we don't want a fraction on the right side (e.g. 1/4). So better to rewrite the first term as:

 4 (y - 1) ²   =   (y - 1) ²   /   (1/4) =  (y - 1) ²  / 0.25

Then we have:

(y - 1) ²   / 0.25     -  (x   - 1) ²   =  1

Or:

  (y - 1) ²   / 0.25   -    (x   - 1) ²  /1     =   1


By way of translation: x' =  x - 1 and y'  =  y - 1

The translated hyperbola equation becomes:

 y'  ²   / 0.25   -    x'  ²  /1     =   1

Which denotes a hyperbola with center at x' = y' = 0 or x = y = 1.

The graph is shown below with asymptotes and foci marked:



The foci F1, F2 are at (0, +  Ö1.25)

The equations for the directrices are: x  =  a/ e and x =   -  a/e

where: a = 0.5   and  e  =   Ö (a ²  +   b ²) /  a  =  Ö1.25 / 0.5 = 2.23

Then: 

x =   0.5/  2.23 =  0.22 and   x = - 0.22