Math / Geometry

QuestionFind the center of the ellipse defined by the equation $\frac{(x+1)^{2}}{9}+\frac{(y+3)^{2}}{16}=1$ . If necessary, round to the nearest tenth.
Answer Attempt 1 out of 2
Center: $\square$ , $\square$

Studdy Solution

STEP 1

1. The equation given is in the standard form of an ellipse.
2. The standard form of an ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ , where $(h, k)$ is the center of the ellipse.

STEP 2

1. Identify the values of $h$ and $k$ from the equation.
2. Determine the center of the ellipse.

STEP 3

Identify the values of $h$ and $k$ from the given equation:
The given equation is:
$\frac{(x+1)^2}{9} + \frac{(y+3)^2}{16} = 1$
Compare this with the standard form:
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
From this comparison, we can see:
$h = -1$ and $k = -3$

STEP 4

Determine the center of the ellipse using the identified values of $h$ and $k$ :
The center of the ellipse is $(h, k) = (-1, -3)$ .
The center of the ellipse is:
$\boxed{-1}, \boxed{-3}$

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