Math / Geometry

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Find the center of the hyperbola defined by the equation $\frac{(y-4)^{2}}{36}-\frac{(x-4)^{2}}{49}=1$ . If necessary, round to the nearest tenth.
Answer Attempt 1 out of 2
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Studdy Solution

STEP 1

1. The given equation is in the standard form of a hyperbola.
2. The equation is $\frac{(y-4)^{2}}{36}-\frac{(x-4)^{2}}{49}=1$ .

STEP 2

1. Identify the standard form of a hyperbola.
2. Determine the center of the hyperbola from the equation.

STEP 3

Identify the standard form of a hyperbola:
The standard form of a hyperbola with a vertical transverse axis is:
$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
where $(h, k)$ is the center of the hyperbola.

STEP 4

Determine the center of the hyperbola from the equation:
Given equation:
$\frac{(y-4)^{2}}{36} - \frac{(x-4)^{2}}{49} = 1$
Compare with the standard form:
$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
From comparison, we identify:
$k = 4$ and $h = 4$
Thus, the center of the hyperbola is $(h, k) = (4, 4)$ .
The center of the hyperbola is:
$\boxed{(4, 4)}$

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