If S and S^{prime} are the foci of the ellips | vedclass

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(D) The given ellipse is $\frac{x^2}{18} + \frac{y^2}{9} = 1$. Here $a^2 = 18$ and $b^2 = 9$,so $a = 3\sqrt{2}$ and $b = 3$.Eccentricity $e = \sqrt{1

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$27$

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(D) The given ellipse is $\frac{x^2}{18} + \frac{y^2}{9} = 1$. Here $a^2 = 18$ and $b^2 = 9$,so $a = 3\sqrt{2}$ and $b = 3$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{18}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.The foci are $S(ae, 0) = (3, 0)$ and $S^{\prime}(-ae, 0) = (-3, 0)$.Let $P = (3\sqrt{2} \cos 	heta, 3 \sin 	heta)$.The focal distances are $SP = a - ex = 3\sqrt{2} - \frac{1}{\sqrt{2}}(3\sqrt{2} \cos 	heta) = 3\sqrt{2} - 3 \cos 	heta$ and $S^{\prime}P = a + ex = 3\sqrt{2} + 3 \cos 	heta$.Then $SP \cdot S^{\prime}P = (3\sqrt{2} - 3 \cos 	heta)(3\sqrt{2} + 3 \cos 	heta) = 18 - 9 \cos^2 	heta$.Since $0 \le \cos^2 	heta \le 1$,the minimum value is $18 - 9(1) = 9$ (at $\cos^2 	heta = 1$) and the maximum value is $18 - 9(0) = 18$ (at $\cos^2 	heta = 0$).Thus,$\min(SP \cdot S^{\prime}P) + \max(SP \cdot S^{\prime}P) = 9 + 18 = 27$.

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and $P$ is a point on the ellipse,then $\min \left(SP \cdot S^{\prime}P\right) + \max \left(SP \cdot S^{\prime}P\right)$ is equal to:

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