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(A) The vertices are $(\pm 6, 0)$,which implies the major axis is along the $x$-axis and $a = 6$.The foci are $(\pm 4, 0)$,which implies $c = 4$.For a

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$\frac{x^2}{36} + \frac{y^2}{20} = 1$

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(A) The vertices are $(\pm 6, 0)$,which implies the major axis is along the $x$-axis and $a = 6$.The foci are $(\pm 4, 0)$,which implies $c = 4$.For an ellipse,the relationship between $a, b,$ and $c$ is $a^2 = b^2 + c^2$.Substituting the values: $6^2 = b^2 + 4^2$.$36 = b^2 + 16$.$b^2 = 36 - 16 = 20$.The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting $a^2 = 36$ and $b^2 = 20$,we get $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6, 0)$,foci $(\pm 4, 0)$.

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