find the standard form of the equation of the ellipse with center (0,0) and foci at (+- 4square root 3, 0) and verticies (+-8,0)
x^2/a^2 + =y^2/b^2 = 1
a = 8
b = sqrt(8^2 - (4sqrt3)^2)
..= sqrt(64 - 48) = sqrt(16) = 4
Thus, x^2/64 + y^2/16 = 1
To find the standard form of the equation of the ellipse with the given information, follow these steps:
1. Identify the center of the ellipse: Given that the center is (0,0), the equation takes the form (x - h)^2 + (y - k)^2 = 1, where (h,k) represents the center.
2. Find the value of 'a': The value of 'a' represents the distance from the center to the vertices of the ellipse. In this case, the vertices are located at (+-8,0). Therefore, a = 8.
3. Determine the value of 'b': The value of 'b' represents the distance from the center to the co-vertices of the ellipse. The co-vertices are located at (0, +-4). In this case, we know that a^2 = 64 and b^2 = a^2 - c^2, where c represents the distance from the center to the foci.
To find c, note that the foci are given as (+-4sqrt(3), 0). Since the center is (0,0), the distance from the center to either of the foci is equal to 4sqrt(3). Therefore, c = 4sqrt(3).
Now, substitute the values of a and c into the equation b^2 = a^2 - c^2:
b^2 = 64 - (4sqrt(3))^2
= 64 - 48
= 16
Taking the square root of both sides, b = 4.
4. Write the equation in standard form: Substitute the values of 'a' and 'b' into the equation (x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h,k) is the center:
(x - 0)^2/64 + (y - 0)^2/16 = 1
Simplifying,
x^2/64 + y^2/16 = 1
Hence, the standard form of the equation for the given ellipse is x^2/64 + y^2/16 = 1.