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Determine the center, transverse axis, vertices, foci, ((y-5)²)/(144)-((x+4)²)/(81)=1

User Boni
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Final answer:

The center of the ellipse is (-4, 5). The transverse axis is parallel to the x-axis and passes through the center. The foci of the ellipse are undefined in the real number system.

Step-by-step explanation:

The given equation is ((y-5)²)/(144)-((x+4)²)/(81)=1. This equation represents an ellipse.

To determine the center of the ellipse, we need to identify the values of h and k in the standard form equation of an ellipse: ((x-h)²)/(a²) + ((y-k)²)/(b²) = 1. Comparing this with the given equation, we can see that h = -4 and k = 5. Therefore, the center of the ellipse is (-4, 5).

The transverse axis of the ellipse is the major axis, which passes through the center and is parallel to the x-axis. The vertices are the points where the ellipse intersects the transverse axis. In this case, the transverse axis has a length of 2a, and a = 9. Therefore, the two vertices are (-4-9, 5) and (-4+9, 5), which simplifies to (-13, 5) and (5, 5).

The foci of the ellipse can be found using the formula c = sqrt(a² - b²), where a = 9 and b = 12. Plugging in these values, we get c = sqrt(81 - 144) = sqrt(-63). Since the square root of a negative number is undefined in the real number system, this ellipse has no real foci.

User Gabriel Osorio
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