Eccentricity of an ellipse.with b > a

Question

I have the following ellipse : $\frac{(x-3)^2}{\frac{9}{4}} + \frac{(y+4)^2}{\frac{25}{4}}=1$

In this case, b > a. It says that to find the eccentricity I must use $\frac{c}{a}$ but I think this is only valid when a>b. What'S the formula to find the eccentricity in my case ?

Answer 1

The formula for eccentricity is always (distance between foci)/(length of major axis).

Answer 2

In this case, $a=\frac 52$ and the major semi-axis is parallel to the $y$ axis, so the eccentricity is still $$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac {9}{25}}=\frac 45$$