If P is a point on the hyperbola 16x^2 - 9y^2 | vedclass

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(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$,which is $\frac{x^2}{3^2

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$6$

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(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$,which is $\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1$.This is a standard hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a = 3$.By the definition of a hyperbola,the absolute difference of the distances of any point $P$ on the hyperbola from the two foci $S_1$ and $S_2$ is equal to the length of the transverse axis,which is $2a$.Therefore,$|PS_1 - PS_2| = 2a = 2(3) = 6$.

If $P$ is a point on the hyperbola $16x^2 - 9y^2 = 144$ whose foci are $S_1$ and $S_2$,then $|PS_1 - PS_2| = $

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