If the latus rectum of a hyperbola is 8 and t | vedclass

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(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$. Given the eccentricity $e = \frac{3}{\sqrt{5}}$,we use t

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$4x^2 - 5y^2 = 100$

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(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$. Given the eccentricity $e = \frac{3}{\sqrt{5}}$,we use the relation $e^2 = 1 + \frac{b^2}{a^2}$. Substituting $e^2 = \frac{9}{5}$ and $b^2 = 4a$,we get $\frac{9}{5} = 1 + \frac{4a}{a^2} = 1 + \frac{4}{a}$. This simplifies to $\frac{4}{5} = \frac{4}{a}$,so $a = 5$. Then $b^2 = 4(5) = 20$. The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which becomes $\frac{x^2}{25} - \frac{y^2}{20} = 1$. Multiplying by $100$,we get $4x^2 - 5y^2 = 100$.

If the latus rectum of a hyperbola is $8$ and the eccentricity is $3/\sqrt{5}$,then the equation of the hyperbola is:

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