The equation is 16x^{2} – 9y^{2} = 576

Let us divide the whole equation by 576, we get

\( \dfrac{16x^2}{576}- \dfrac{9y^2}{576}\)

= \( \dfrac{576}{576}\)

\( \dfrac{x^2}{36}- \dfrac{y^2}{64}=1\)

On comparing this equation with the standard equation of hyperbola

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

We get a = 6 and b = 8,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 36 + 64

c^{2} = \( \sqrt{100}\)

c = 10

Then,

The coordinates of the foci are (10, 0) and (-10, 0).

The coordinates of the vertices are (6, 0) and (-6, 0).

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

= \( \dfrac{10}{6}\) =\( \dfrac{5}{3}\)

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

= \( \dfrac{(2×8^2)}{6}\)

= \( \dfrac{(2×64)}{6}\)

= \( \dfrac{64}{3}\)

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