Find the equation of the hyperbola satisfying | vedclass

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(A) The foci are $(\pm 5, 0)$,which lie on the $x$-axis. The standard form of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Given t

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$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

Vedclass · Answer

(A) The foci are $(\pm 5, 0)$,which lie on the $x$-axis. The standard form of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Given the foci $(\pm c, 0) = (\pm 5, 0)$,we have $c = 5$.The length of the transverse axis is $2a = 8$,which gives $a = 4$.Using the relation $c^{2} = a^{2} + b^{2}$,we get $5^{2} = 4^{2} + b^{2}$.$25 = 16 + b^{2} \Rightarrow b^{2} = 25 - 16 = 9$.Substituting $a^{2} = 16$ and $b^{2} = 9$ into the standard equation,we get $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 5, 0)$,the transverse axis is of length $8$.

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