find the equation of the hyperbola satisfying the given condition:

vertices (± 7, 0), \[e = \frac{4}{3}\]

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#### Solution

The vertices of hyperbola are \[\left( \pm 7, 0 \right)\] and eccentricity is \[\frac{4}{3}\] Thus, the value of \[a = 7\].

Now, using the relation

\[b^2 = a^2 ( e^2 - 1)\], we get:

\[\Rightarrow b^2 = 49\left( \frac{16}{9} - 1 \right)\]

\[ \Rightarrow b^2 = 49 \times \frac{7}{9} = \frac{343}{9}\]

Thus, the equation of the hyperbola is

\[\frac{x^2}{49} - \frac{9 y^2}{343} = 1\] .

Is there an error in this question or solution?

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