Final answer:
The given equation represents an ellipse with center (3, 4). The foci are located at (3, 4 ± √6), and the vertices are at (3 ± √24, 4). The focal axis is y=4, and the conjugate axis is x=3.
Step-by-step explanation:
To identify the conic section represented by the equation 8y2+6x2−36x−64y+38=0, we need to rewrite it in a standard form. This involves completing the square for both x and y terms. Given that we have both x2 and y2 terms, the conic section is likely an ellipse or a hyperbola, but not a parabola (since parabolas have only one squared term).
Grouping the x and y terms and completing the square gives us:
6(x2 - 6x) + 8(y2 - 8y) + 38 = 0
6[(x - 3)2 - 9] + 8[(y - 4)2 - 16] + 38 = 0
6(x - 3)2 + 8(y - 4)2 - 54 - 128 + 38 = 0
6(x - 3)2 + 8(y - 4)2 = 144
Dividing by 144 to get the equation in standard form:
(x - 3)2/24 + (y - 4)2/18 = 1
This is the standard form of an ellipse equation, which is (x - h)2/a2 + (y - k)2/b2 = 1, where (h, k) is the center of the ellipse.
Thus, the center of the ellipse is at (3, 4). The lengths of the semi-major axis a and semi-minor axis b are √24 and √18, respectively. To find the foci, we use the formula c=√(a2-b2), and since 24 > 18, a=√24 and b=√18. Hence, c=√(24-18)=√6.
The foci of the ellipse are therefore at (3, 4 ± √6). The vertices of the ellipse are found at the endpoints of the major axis, so they are at (3 ± √24, 4).
The focal axis is the major axis, with equation y=4. The conjugate axis is the minor axis, with equation x=3.