The equation is \( \dfrac{x^2}{16}\) + \( \dfrac{y^2}{9}\)= 1 or

\( \dfrac{x^2}{4^2}\) + \( \dfrac{y^2}{3^2}\)= 1

Here, the denominator of \( \dfrac{x^2}{16}\) is greater than the denominator of \( \dfrac{y^2}{9}\).

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with

\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1, we get

a = 4 and b = 3.

c = \( \sqrt{ (a^2 – b^2)}\)

= \( \sqrt{ (16– 9)}\)

= \( \sqrt{7}\)

Then,

The coordinates of the foci are (\( \sqrt{7}\), 0) and (\(- \sqrt{7}\), 0).

The coordinates of the vertices are (4, 0) and (-4, 0)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = \( \dfrac{c}{a}\) = \( \dfrac{\sqrt{7}}{4}\)

Length of latus rectum =\( \dfrac{2b^2}{a}\)

= \( \dfrac{(2×3^2)}{4}\) = \( \dfrac{(2×9)}{4}\)

= \( \dfrac{18}{4}\)

= \( \dfrac{9}{2}\)

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