What is the standard equation of a hyperbola given the vertices (7,-2) , (5,-2) and asymptotes " y= 11x - 68 " & " y= 11x + 64 "?
center is at (6,-2)
a = 1
b/a = 11, so b=11
To find the standard equation of a hyperbola given the vertices and asymptotes, we need to determine the center, the distance between the center and the vertices, and the slope of the asymptotes.
1. Find the center:
The center of a hyperbola is the midpoint between the vertices. The x-coordinate of the center is the average of the x-coordinates of the vertices, and the y-coordinate of the center is the average of the y-coordinates.
x-coordinate of the center = (7 + 5) / 2 = 6
y-coordinate of the center = (-2 - 2) / 2 = -2
Therefore, the center of the hyperbola is (6, -2).
2. Find the distance between the center and the vertices:
The distance between the center and the vertices is called the semi-major axis. It is half the distance between the x-coordinates of the vertices.
Semi-major axis = (7 - 5) / 2 = 1
3. Find the slope of the asymptotes:
The given equations of the asymptotes, y = 11x - 68 and y = 11x + 64, are in the form y = mx + b, where m is the slope of the line. In this case, the slopes of the asymptotes are both 11.
4. Determine if the transverse axis is horizontal or vertical:
Since the given vertices have the same y-coordinate, the transverse axis is horizontal.
Now we can write the standard equation of the hyperbola. The general equation for a hyperbola with a horizontal transverse axis and center (h, k) is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1,
where (h, k) represents the center of the hyperbola, a is the semi-major axis, and b is the semi-minor axis.
By substituting the values we found, the equation becomes:
(x - 6)^2 / 1^2 - (y + 2)^2 / b^2 = 1.
Since we still need to determine the value of b, we'll use the slope of the asymptotes to find it.
The slope of the asymptotes is given by b / a, so in this case:
11 = b / 1.
Therefore, b = 11.
Now we can substitute the value of b into our equation:
(x - 6)^2 - (y + 2)^2 / 11^2 = 1.
Simplifying further:
(x - 6)^2 - (y + 2)^2 / 121 = 1.
Therefore, the standard equation of the hyperbola is:
(x - 6)^2 - (y + 2)^2 / 121 = 1.
To find the standard equation of a hyperbola, you need to know the coordinates of the vertices and the equations of the asymptotes. Given that the vertices of the hyperbola are (7, -2) and (5, -2), and the asymptotes are y = 11x - 68 and y = 11x + 64, we can proceed as follows:
Step 1: Determine the center of the hyperbola.
The center of the hyperbola is the midpoint between the vertices. To find the midpoint, add the x-coordinates and divide by 2, and add the y-coordinates and divide by 2.
Center = ( (7+5)/2 , (-2-2)/2 ) = (6, -2)
Step 2: Determine the distance between the center and one of the vertices.
The distance between the center and a vertex is the distance between the x-coordinates or the distance between the y-coordinates, whichever is non-zero.
Distance between center and vertex = |7-6| = 1
Step 3: Determine the equations of the asymptotes in slope-intercept form.
Given that the equations of the asymptotes are y = 11x - 68 and y = 11x + 64, we can rewrite them in the slope-intercept form (y = mx + b), where "m" is the slope and "b" is the y-intercept.
Asymptote 1: y = 11x - 68
Asymptote 2: y = 11x + 64
Step 4: Determine the distances between the center and the vertices along the transverse axis and the conjugate axis.
Since the asymptotes have a slope of 11 (which is the coefficient of x), the distance between the center and a vertex along the transverse axis is the distance between the x-coordinates divided by the absolute value of the coefficient of x. The distance between the center and a vertex along the conjugate axis is the distance between the y-coordinates divided by the absolute value of the coefficient of y.
Distance between center and vertex along transverse axis = 1 / |11| = 1/11
Distance between center and vertex along conjugate axis = 0 (since the y-coordinates are the same for both vertices)
Step 5: Determine the major and minor axes.
The major axis is the axis along which the two vertices lie. In this case, the major axis is parallel to the x-axis, since the y-coordinates of the vertices are the same. The length of the major axis is twice the distance between the center and a vertex along the transverse axis.
Major axis = 2 * (1/11) = 2/11
The minor axis is perpendicular to the major axis and passes through the center. In this case, the length of the minor axis is twice the distance between the center and a vertex along the conjugate axis.
Minor axis = 2 * 0 = 0
Step 6: Determine the eccentricity of the hyperbola.
The eccentricity of a hyperbola is given by the ratio of the distance between the center and a vertex along the transverse axis to the distance between the center and a vertex along the conjugate axis.
Eccentricity (e) = (1/11) / 0 = undefined
Since the eccentricity is undefined (greater than 1), we know that the hyperbola is an "oblique" hyperbola.
Step 7: Write the standard equation of the hyperbola.
The standard equation of a hyperbola with the center (h, k), the semi-major axis "a", the semi-minor axis "b", and the slope of the transverse axis "m" is given by:
[(x-h)^2 / a^2] - [(y-k)^2 / b^2] = 1
In our case:
Center = (6, -2)
Semi-major axis (a) = 1/11 (half the length of the major axis)
Semi-minor axis (b) = 0 (since the length of the minor axis is zero)
Since the hyperbola is oblique, we need the slope of the transverse axis. We can calculate the slope from the asymptotes. The slope of the transverse axis is the negative reciprocal of the slope of the asymptotes.
Slope of the asymptote = 11
Slope of the transverse axis = -1/11
Substituting these values into the formula:
[(x-6)^2 / (1/11)^2] - [(y+2)^2 / 0^2] = 1
Simplifying further:
11(x-6)^2 = 1
Therefore, the standard equation of the hyperbola, given the vertices (7, -2) and (5, -2), and asymptotes y = 11x - 68 and y = 11x + 64, is:
11(x-6)^2 = 1