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
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
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