The latus-rectum of the hyperbola 16x^2 - 9y^ | vedclass

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(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$. Dividing both sides by $144$,we get: $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Comparing

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$\frac{32}{3}$

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(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$. Dividing both sides by $144$,we get: $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 16$. Thus,$a = 3$ and $b = 4$. The length of the latus-rectum is given by the formula $\frac{2b^2}{a}$. Substituting the values,we get: $L.R. = \frac{2 	imes 16}{3} = \frac{32}{3}$.

The latus-rectum of the hyperbola $16x^2 - 9y^2 = 144$ is

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