# 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), $e = \frac{4}{3}$

#### 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

$\frac{x^2}{49} - \frac{9 y^2}{343} = 1$ .
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#### APPEARS IN

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