If S is the focus of the ellipse frac{x^2}{9} | vedclass

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(B) For the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$,we have $a^2=9$ and $b^2=4$. Thus,$a=3$ and $b=2$. The eccentricity $e$ is given by $e = \sqrt{1 -

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$\frac{2}{\sqrt{5}}$

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(B) For the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$,we have $a^2=9$ and $b^2=4$. Thus,$a=3$ and $b=2$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. The focus $S$ on the positive $X$-axis is $(ae, 0) = (3 	imes \frac{\sqrt{5}}{3}, 0) = (\sqrt{5}, 0)$. $A$ point $P$ on the ellipse is $(3 \cos 	heta, 2 \sin 	heta)$. The distance $SP$ is given by the focal distance formula $SP = a - ex$ for a focus on the positive $X$-axis,where $x = 3 \cos 	heta$. So,$SP = 3 - (\frac{\sqrt{5}}{3})(3 \cos 	heta) = 3 - \sqrt{5} \cos 	heta$. Given $SP = 1$,we have $1 = 3 - \sqrt{5} \cos 	heta$. $\sqrt{5} \cos 	heta = 2$. $\cos 	heta = \frac{2}{\sqrt{5}}$.

If $S$ is the focus of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ lying on the positive $X$-axis and $P(\theta)$ is a point on the ellipse such that $SP=1$,then $\cos \theta=$

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