P(theta) is a point on the hyperbola frac{x^2 | vedclass

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(C) The focus $S$ is $(ae, 0)$. Given $Q = (0, 1)$,$S Q^2 = (ae)^2 + 1^2 = 26$,so $a^2 e^2 = 25$,which means $ae = 5$.For the hyperbola $\frac{x^2}{a^

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$\frac{\pi}{3}$

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(C) The focus $S$ is $(ae, 0)$. Given $Q = (0, 1)$,$S Q^2 = (ae)^2 + 1^2 = 26$,so $a^2 e^2 = 25$,which means $ae = 5$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$,we have $b^2 = 9$,so $e^2 = 1 + \frac{9}{a^2}$.Substituting $e^2 = \frac{25}{a^2}$,we get $\frac{25}{a^2} = 1 + \frac{9}{a^2}$,which implies $\frac{16}{a^2} = 1$,so $a = 4$.Then $e = \frac{5}{4}$. The point $P$ is $(4 \sec 	heta, 3 	an 	heta)$.The distance $SP = 6$. Using the focal distance formula $SP = |ae \sec 	heta - a|$ is not directly applicable as $S$ is $(ae, 0)$,so $SP^2 = (4 \sec 	heta - 5)^2 + (3 	an 	heta - 0)^2 = 36$.$16 \sec^2 	heta - 40 \sec 	heta + 25 + 9 	an^2 	heta = 36$.Using $	an^2 	heta = \sec^2 	heta - 1$,we get $16 \sec^2 	heta - 40 \sec 	heta + 25 + 9 \sec^2 	heta - 9 = 36$.$25 \sec^2 	heta - 40 \sec 	heta - 20 = 0$,which simplifies to $5 \sec^2 	heta - 8 \sec 	heta - 4 = 0$.$(5 \sec 	heta + 2)(\sec 	heta - 2) = 0$.Since $|\sec 	heta| \geq 1$,we have $\sec 	heta = 2$,which gives $	heta = \frac{\pi}{3}$.

$P(\theta)$ is a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$,$S$ is its focus lying on the positive $X$-axis and $Q = (0, 1)$. If $S Q = \sqrt{26}$ and $S P = 6$,then $\theta =$

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