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Find the Equation of the Hyperbola Satisfying the Given Condition: Vertices (± 7, 0), E = 4 3 - Mathematics

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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 11.08 | Page 14
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