Find the equation of the hyperbola satisfying | vedclass

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Question

(A) The vertices are $(\pm 7, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given $a = 7$ a

Vedclass · Accepted Answer

$\frac{x^2}{49} - \frac{9y^2}{343} = 1$

Vedclass · Answer

(A) The vertices are $(\pm 7, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given $a = 7$ and eccentricity $e = \frac{c}{a} = \frac{4}{3}$. Substituting $a = 7$,we get $\frac{c}{7} = \frac{4}{3}$,so $c = \frac{28}{3}$. Using the relation $c^2 = a^2 + b^2$,we have $b^2 = c^2 - a^2 = (\frac{28}{3})^2 - 7^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}$. Substituting $a^2 = 49$ and $b^2 = \frac{343}{9}$ into the standard equation,we get $\frac{x^2}{49} - \frac{y^2}{343/9} = 1$,which simplifies to $\frac{x^2}{49} - \frac{9y^2}{343} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 7, 0)$,$e = \frac{4}{3}$.

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