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 is 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
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.
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
(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
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
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