Let S and S^{prime} be the foci of the ellips | vedclass

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(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$, we have $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} =

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$13$

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(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$, we have $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$. The foci are $S(ae, 0) = (4, 0)$ and $S^{\prime}(-ae, 0) = (-4, 0)$. Since $P(\alpha, \beta)$ lies on the ellipse, $SP+S^{\prime}P=2a=10$. Given $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$. Using $(SP+S^{\prime}P)^2 = (SP)^2+(S^{\prime}P)^2+2SP \cdot S^{\prime}P$, we substitute $(SP)^2+(S^{\prime}P)^2 = (SP+S^{\prime}P)^2-2SP \cdot S^{\prime}P$. So, $(SP+S^{\prime}P)^2-3SP \cdot S^{\prime}P=37$. $10^2-3SP \cdot S^{\prime}P=37$ $\Rightarrow 100-3SP \cdot S^{\prime}P=37$ $\Rightarrow 3SP \cdot S^{\prime}P=63$ $\Rightarrow SP \cdot S^{\prime}P=21$. For any point $P(\alpha, \beta)$ on the ellipse, $SP = a-e\alpha$ and $S^{\prime}P = a+e\alpha$. $SP \cdot S^{\prime}P = a^2-e^2\alpha^2 = 25-\frac{16}{25}\alpha^2 = 21$. $\frac{16}{25}\alpha^2 = 4 \Rightarrow \alpha^2 = \frac{100}{16} = \frac{25}{4}$. From $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$, we have $\frac{25/4}{25}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{1}{4}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{\beta^2}{9}=\frac{3}{4}$ $\Rightarrow \beta^2=\frac{27}{4}$. Thus, $\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4} = \frac{52}{4} = 13$.

Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$, then $\alpha^2+\beta^2$ is equal to:

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