Question

Determine the eccentricity of the ellipse given by 9x^2+16y^2-72x+64y-368=0

Answer 1

Answer:

$e=0.66$

Step-by-step explanation:

The eccentricity of an ellipse is given by: $e=\sqrt{1-\frac{b^2}{a^2} }$

The given ellipse has equation: $9x^2+16y^2-72x+64y-368=0$

We can rewrite this equation in standard form to obtain:

$\frac{(x-4)^2}{8^2}+\frac{y+2}{6^2}=1$

We compare to the general standard form equation: $\frac{(x-h)^2}{a^2}+\frac{y-k}{b^2}=1$

to get: $a=8,b=6$

We substitute into the eccentricity formula to get:

$e=\sqrt{1-\frac{6^2}{8^2} }$

$e=\frac{\sqrt{7}}{4}=0.661438$

The eccentricity is 0.66 to the nearest hundredth