What is the equation of hyperbola whose foci at F(-8,0) and length of trasverse axis is 6?
Assuming the other focus is at (8,0) then
c^2 = a^2+b^2, so
84 = 36+b^2
b^2 = 48
x^2/64 - y^2/48 = 1
To find the equation of a hyperbola given the foci and the length of the transverse axis, you can use the standard form of the hyperbola 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 represent the lengths of the semi-major and semi-minor axes, respectively.
In this case, the foci are given as F(-8, 0), which means the center of the hyperbola is at the point (-8, 0).
The length of the transverse axis is given as 6, which means the distance between the two vertices (endpoints of the transverse axis) is 6 units.
Since a hyperbola is symmetrical, the distance from the center to each vertex is half of the transverse axis length. Therefore, a = 6/2 = 3.
To find the value of b, you can use the relationship between a, b, and c (the distance from the center to the foci). In a hyperbola, c is related to a and b by the equation c^2 = a^2 + b^2.
Since the distance between the foci is 2a, we have:
c = 2a = 2(3) = 6
Now we can substitute the values into the equation to get the final equation of the hyperbola:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
(x - (-8))^2 / 3^2 - (y - 0)^2 / b^2 = 1
(x + 8)^2 / 9 - (y - 0)^2 / b^2 = 1
Thus, the equation of the hyperbola is (x + 8)^2 / 9 - (y - 0)^2 / b^2 = 1, where b is the length of the semi-minor axis, which can be found using c^2 = a^2 + b^2.