Find the equation of hyperbola whose foci are at(10,2) (0,2)and whose eccentricity is10/3
The center of the graph is at (5,2), so the equation will be
(x-5)^2/a^2 - (y-2)^2/b^2 = 1
Now, we have
c = 5
e = c/a = 10/3, so a = 3/2
a^2+b^2=c^2, so
9/4 + b^2 = 100/4, so b^2 = 91/4
The equation is thus
(x-5)^2/(9/4) - (y-2)^2/(91/4) = 1
or
4(x-5)^2/9 - 4(y-2)^2/91 = 1
Confirm this at
http://www.wolframalpha.com/input/?i=hyperbola+%28x-5%29^2%2F%289%2F4%29+-+%28y-2%29^2%2F%2891%2F4%29+%3D+1
To find the equation of a hyperbola given its foci and eccentricity, we can use the following standard form equation:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1
Where (h, k) represents the center of the hyperbola, and 'a' and 'b' are the lengths of the semi-major and semi-minor axes, respectively.
1. Find the center of the hyperbola:
The center of the hyperbola is the midpoint between the two foci. In this case, the foci are at (10, 2) and (0, 2).
=> The x-coordinate of the center = (10 + 0) / 2 = 5
=> The y-coordinate of the center = (2 + 2) / 2 = 2
=> The center of the hyperbola is (5, 2).
2. Find the distance between the foci:
The distance between the foci can be calculated using the eccentricity (e) by the formula:
c = a * e, where 'c' is the distance between the foci.
In this case, the eccentricity (e) = 10/3.
=> c = a * (10/3)
3. Determine the length of the semi-major axis (a):
To find 'a', we can make use of the distance formula between the two foci ((x1, y1) and (x2, y2)):
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, d = c = a * (10/3),
=> a * (10/3) = sqrt((10 - 0)^2 + (2 - 2)^2)
=> a * (10/3) = sqrt(100)
=> a * (10/3) = 10
=> a = 3
4. Find the length of the semi-minor axis (b):
The eccentricity (e) can also be expressed as:
e = sqrt(a^2 + b^2) / a
Given that e = (10/3) and a = 3, we can solve for 'b':
(10/3) = sqrt(3^2 + b^2) / 3
10 = sqrt(9 + b^2)
100 = 9 + b^2
b^2 = 100 - 9
b^2 = 91
b = sqrt(91)
Now we have all the required information to write the equation of the hyperbola:
[(x - 5)^2 / 3^2] - [(y - 2)^2 / (√91)^2] = 1
Simplifying further, we have:
[(x - 5)^2 / 9] - [(y - 2)^2 / 91] = 1
Therefore, the equation of the hyperbola with foci at (10, 2) and (0, 2), and eccentricity 10/3 is:
[(x - 5)^2 / 9] - [(y - 2)^2 / 91] = 1.