Question

question 2 of 10
what is the eccentricity of the ellipse shown below?
\\(\frac{(x - 5)^2}{52}+\frac{(y + 1)^2}{64}=1\\)
a. \\(\frac{2}{\sqrt{3}}\\)
b. \\(\sqrt{3}\\)
c. \\(\frac{\sqrt{3}}{2}\\)
d. \\(\frac{\sqrt{3}}{4}\\)

Explanation:

Step1: Identify $a$ and $b$ values

For the ellipse equation $\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1$ (since $a^2>b^2$), here $a^2 = 64$, so $a = 8$, and $b^2=52$, so $b = 2\sqrt{13}$.

Step2: Use the eccentricity formula

The formula for the eccentricity $e$ of an ellipse is $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$. Substitute $a = 8$ and $b = 2\sqrt{13}$ into it: $e=\sqrt{1-\frac{52}{64}}=\sqrt{\frac{64 - 52}{64}}=\sqrt{\frac{12}{64}}=\frac{\sqrt{12}}{8}=\frac{2\sqrt{3}}{8}=\frac{\sqrt{3}}{4}$.

Answer:

D. $\frac{\sqrt{3}}{4}$