Let S be the focus of the hyperbola frac{x^2} | vedclass

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(C) Given the hyperbola equation $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. Since point $P(5, y_1)$ lies on the hyperbola,we subst

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$\frac{9}{4}$

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(C) Given the hyperbola equation $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. Since point $P(5, y_1)$ lies on the hyperbola,we substitute $x=5$: $\frac{25}{16}-\frac{y_1^2}{9}=1$ $\Rightarrow \frac{y_1^2}{9}=\frac{25}{16}-1 = \frac{9}{16}$ $\Rightarrow y_1^2 = \frac{81}{16}$ $\Rightarrow y_1 = \pm \frac{9}{4}$. Thus,$P = (5, \pm \frac{9}{4})$. The eccentricity $e$ is given by $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$. The focus $S$ on the positive $X$-axis is $(ae, 0) = (4 	imes \frac{5}{4}, 0) = (5, 0)$. Now,the distance $SP$ is: $SP = \sqrt{(5-5)^2 + (0 - (\pm \frac{9}{4}))^2} = \sqrt{0 + \frac{81}{16}} = \frac{9}{4}$.

Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X$-axis and $P(5, y_1)$ be a point on the hyperbola. Then $SP =$

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