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
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
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