what is the foci of the graph x^2/36-y^2/16=1
To find the foci of the graph of the equation x^2/36 - y^2/16 = 1, you can use the formula for the foci of a hyperbola.
The standard form of a hyperbola with a horizontal transverse axis is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
In this case, we have:
x^2 / 36 - y^2 / 16 = 1
Comparing these equations, we can see that h = 0, k = 0, a^2 = 36, and b^2 = 16.
To find the foci, we will use the formula:
c^2 = a^2 + b^2
where c represents the distance from the center to the foci.
Plugging in the values, we get:
c^2 = 36 + 16
c^2 = 52
Taking the square root of both sides, we find:
c ≈ √52 ≈ 7.211
Since the center of the hyperbola is at (h, k), which is (0, 0) in this case, the foci will be located at (0 ± c, 0).
Therefore, the foci of the graph are approximately (7.211, 0) and (-7.211, 0).
To determine the foci of the graph of the equation x^2/36 - y^2/16 = 1, you need to use the standard equation for a hyperbola. The standard equation for a horizontal hyperbola is (x - h)^2/a^2 - (y - k)^2/b^2 = 1, where (h, k) represents the center of the hyperbola.
Comparing this standard equation with the given equation x^2/36 - y^2/16 = 1, we can observe that h = 0, k = 0, a = 6, and b = 4. Therefore, the center of the hyperbola is at (0, 0).
The formula to find the foci of a hyperbola with a horizontal major axis is given by f = √(a^2 + b^2), where f represents the distance between the center and the foci.
In this case, we have a = 6 and b = 4, so plugging these values into the formula gives us:
f = √(6^2 + 4^2) = √(36 + 16) = √52 ≈ 7.211.
Therefore, the foci of the graph x^2/36 - y^2/16 = 1 are located at a distance of approximately 7.211 units from the center along the horizontal axis in both directions.
That is the equation of a hyperbola.
(x/6)^2 - (y/4)^2 = 1
= (x/a)^2 - (y/b)^2
where a = 6 and b = 4.
The foci are at (-c, 0) and (c,0), where
c = sqrt (a^2 + b^2) = sqrt 52 = 7.21