HELP! I am not sure how I would even start this problem or solve it.
A hyperbola with a horizontal transverse axis contains the point at (4, 3). The equations of the asymptotes are y-x=1 and y+x=5 Write an equation for the hyperbola.
Sketch those lines slopes of +/- 1
center axis halfway between 1 and 5 so at y = 3
the lines cross at (2,3), so that is center of hyperbola
so
(x-2)^2/a^2 - (y-3)^2/b^2 = 1
slopes of asymptotes = +/- b/a so b = a
(x-2)^2/a^2 - (y-3)^2/a^2 = 1
put that point in now
(4-2)^2/a^2 - (3-3)^2/a^2 = 1
a^2 = 4
a = 2
(x-2)^2/4 - (y-3)^2/4 = 1
To write the equation of the hyperbola, you need to determine some key information about it. Here are the steps you can follow:
Step 1: Identify the center of the hyperbola.
In this case, the hyperbola has a horizontal transverse axis, so the center will have the form (h, k). Since the asymptotes intersect at the point (2,3) and (4,5), the center of the hyperbola will be the midpoint of these two points. Let's calculate the midpoint:
Midpoint x-coordinate = (2 + 4)/2 = 3
Midpoint y-coordinate = (3 + 5)/2 = 4
So, the center of the hyperbola is (3, 4).
Step 2: Determine the distance between the center and the vertices.
For a hyperbola, the distance between the center and the vertices along the transverse axis is called the "transverse axis length." In this case, the transverse axis is horizontal, and the asymptotes are parallel to the transverse axis. The transverse axis length is equal to half the distance between the intersection points of the asymptotes.
Distance between (4,3) and (3,4) = sqrt[(4-3)^2 + (3-4)^2] = sqrt[1 + 1] = sqrt(2)
So, the transverse axis length is sqrt(2).
Step 3: Determine the distance between the center and the foci.
For a hyperbola, the distance between the center and the foci along the transverse axis is called the "focal length." The focal length can be calculated using the following formula:
Focal Length (c) = sqrt(a^2 + b^2), where a is half the length of the transverse axis, and b is half the length of the conjugate axis.
In this case, the transverse axis length is sqrt(2), and the asymptotes intersect at right angles, so the conjugate axis length is also sqrt(2). Therefore:
a = sqrt(2)/2
b = sqrt(2)/2
Plugging the values of a and b into the formula:
Focal Length (c) = sqrt[(sqrt(2)/2)^2 + (sqrt(2)/2)^2] = sqrt[(2/4) + (2/4)] = sqrt(1) = 1
So, the focal length is 1.
Step 4: Determine the orientation of the hyperbola.
Since the equations of the asymptotes are y-x=1 and y+x=5, we can compare these to the standard form of the equations of asymptotes:
y - x = ± (a/b)(x - h) + k
y + x = ± (a/b)(x - h) + k
Comparing these equations to the given asymptotes, we can see that the slopes and intercepts are the same, except for positive/negative correlations. This indicates that the hyperbola is oriented diagonally.
Step 5: Write the equation of the hyperbola.
The standard form of a hyperbola equation for a horizontally oriented hyperbola with the center at (h, k) is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
With the information we've obtained, the equation of the hyperbola is:
(x - 3)^2 / (sqrt(2)/2)^2 - (y - 4)^2 / (sqrt(2)/2)^2 = 1
Simplifying the equation gives:
(x - 3)^2 / (2/4) - (y - 4)^2 / (2/4) = 1
(x - 3)^2 / (1/2) - (y - 4)^2 / (1/2) = 1
2(x - 3)^2 - 2(y - 4)^2 = 1
Therefore, the equation of the hyperbola is 2(x - 3)^2 - 2(y - 4)^2 = 1.