Given:

Vertices (±7, 0) and e = \( \dfrac{4}{3}\)

Here, the vertices are on the x- axis

The equation of the hyperbola is of the form

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

Since, the vertices are (± 7, 0), so, a = 7

It is given that e = \( \dfrac{4}{3}\)

\( \dfrac{c}{a}\) = \( \dfrac{4}{3}\)

3c = 4a

Substitute the value of a, we get

3c = 4(7)

c = \( \dfrac{28}{3}\)

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

7^{2} + b^{2} = (\( \dfrac{28}{3}\))^{2}

b^{2} = \( \dfrac{784}{9}\) – 49

= \( \dfrac{(784 – 441)}{9}\)

= \( \dfrac{343}{9}\)

The equation of the hyperbola is

\( \dfrac{x^2}{49}- \dfrac{9y^2}{343}=1\)

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