Question

: What is the eccentricity of the ellipse shown below?

(x-5)^2 (y+1)^2
--------- + ----------- = 1
52 64

A: √3
B: 2/√3
C: √3/4
D: √3/2

Answer 1

Answer on Question #57274 – Math – Analytic Geometry

Question

What is the eccentricity of the ellipse shown below?

$$\frac{(x - 5)^2}{52} + \frac{(y + 1)^2}{64} = 1$$

A: $\sqrt{3}$

B: $\frac{2}{\sqrt{3}}$

C: $\frac{\sqrt{3}}{4}$

D: $\frac{\sqrt{3}}{2}$

Solution

If the equation of the ellipse is

$$\frac{(x - x_0)^2}{b^2} + \frac{(y - y_0)^2}{a^2} = 1,$$

then in equation

$$\frac{(x - 5)^2}{52} + \frac{(y + 1)^2}{64} = 1$$

$a^2 = 64$, hence $a = 8$; $b^2 = 52$. The ellipse is taller-than-wide in this case.

Next, $c^2 = a^2 - b^2$, hence $c = 2\sqrt{3}$.

The eccentricity is given by

$$\varepsilon = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a}, \text{ hence } \varepsilon = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4}.$$

Answer: C: $\frac{\sqrt{3}}{4}$.

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