Advertisements
Advertisements
Question
find the equation of the hyperbola satisfying the given condition:
vertices (± 7, 0), \[e = \frac{4}{3}\]
Advertisements
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
\[\frac{x^2}{49} - \frac{9 y^2}{343} = 1\] .
shaalaa.com
Is there an error in this question or solution?
