Q.For the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, find $a$, $b$, and the eccentricity $e$.

Reading the two denominators off the standard ellipse equation gives $a^2$ and $b^2$ directly; from these, $c$ and then the eccentricity follow using the ellipse relation $c^2=a^2-b^2$.

Comparing $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ with the standard ellipse form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (with $a>b$, since $25>16$):

For an ellipse, $c^2=a^2-b^2$:

Eccentricity: $e=\dfrac{c}{a}=\dfrac35=0.6$.

Using $c^2=a^2+b^2$ (the hyperbola relation) instead of $c^2=a^2-b^2$ for an ellipse is the single most common error in this topic — always double-check which conic's relation is being applied. Also confirm …