To determine the eccentricity of the ellipse given by the equation [tex]\(9 x^2 + 16 y^2 - 72 x + 64 y - 368 = 0\)[/tex], we can follow these steps:
1. Rewrite the equation in standard form by completing the square:
Starting with the given equation:
[tex]\[
9x^2 + 16y^2 - 72x + 64y - 368 = 0
\][/tex]
We group the terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
9(x^2 - 8x) + 16(y^2 + 4y) = 368
\][/tex]
2. Complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
For the [tex]\(x\)[/tex]-terms:
[tex]\[
x^2 - 8x = (x - 4)^2 - 16
\][/tex]
For the [tex]\(y\)[/tex]-terms:
[tex]\[
y^2 + 4y = (y + 2)^2 - 4
\][/tex]
Substituting these back into the equation:
[tex]\[
9((x - 4)^2 - 16) + 16((y + 2)^2 - 4) = 368
\][/tex]
Simplify the equation:
[tex]\[
9(x - 4)^2 - 144 + 16(y + 2)^2 - 64 = 368
\][/tex]
Combine constants on the right-hand side:
[tex]\[
9(x - 4)^2 + 16(y + 2)^2 = 576
\][/tex]
3. Express the equation in the standard form of an ellipse:
Divide the entire equation by 576 to normalize it:
[tex]\[
\frac{9(x - 4)^2}{576} + \frac{16(y + 2)^2}{576} = 1
\][/tex]
This simplifies to:
[tex]\[
\frac{(x - 4)^2}{64} + \frac{(y + 2)^2}{36} = 1
\][/tex]
This is the standard form of the ellipse equation, [tex]\(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\)[/tex], where [tex]\(a^2 = 64\)[/tex] and [tex]\(b^2 = 36\)[/tex].
4. Identify the lengths of the semi-major and semi-minor axes:
[tex]\[
a = \sqrt{64} = 8
\][/tex]
[tex]\[
b = \sqrt{36} = 6
\][/tex]
5. Calculate the eccentricity from [tex]\(c^2 = a^2 - b^2\)[/tex]:
[tex]\[
c^2 = a^2 - b^2 = 64 - 36 = 28
\][/tex]
[tex]\[
c = \sqrt{28} \approx 5.291502622129181
\][/tex]
The eccentricity [tex]\(e\)[/tex] is given by:
[tex]\[
e = \frac{c}{a} = \frac{5.291502622129181}{8} \approx 0.661437827766123
\][/tex]
6. Round the eccentricity to two decimal places:
[tex]\[
e \approx 0.66
\][/tex]
Therefore, the eccentricity of the given ellipse is [tex]\(0.66\)[/tex]. The correct answer is:
[tex]\(\boxed{0.66}\)[/tex]
