Q.Find the coordinates of the foci, the vertices, the eccentricity, the lengths of the major and minor axes, and the length of the latus rectum of the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$.

Using the focal-distance definition of the ellipse and the point at the end of the minor axis, $a^2=b^2+c^2$ with $c>0$ forces $a>b$.

Let the ellipse have foci $S(c,0)$ and $S'(-c,0)$ (with $c>0$, since a genuine ellipse — not a circle — has two distinct foci), and let the constant sum of focal distances be $2a$ (this defines the semi-major axis length $a$, so major axis $=2a$).

Let $B=(0,b)$ be the point where the ellipse meets the $y$-axis (the end of what will be called the minor axis, length $2b$). By symmetry about the $y$-axis, $B$ is equidistant from both foci:

$BS=BS'=\sqrt{b^2+c^2}.$

By the defining property of the ellipse, $BS+BS'=2a$, so:

$2\sqrt{b^2+c^2}=2a\implies\sqrt{b^2+c^2}=a\implies a^2=b^2+c^2.$

Since the foci are genuinely distinct from the centre (a real ellipse, not a circle), $c>0$, so $c^2>0$. Hence:

$a^2=b^2+c^2>b^2\implies a>b\ (\text{since }a,b>0).$

Therefore the major axis ($2a$) is greater than the minor axis ($2b$).