Question
Let $S$ and $S'$ 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 $(SP)^{2} + \left( S'P \right)^{2} - SP \bullet S'P = 37$, then $\alpha^{2} + \beta^{2}$ is equal to :
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Solution
$\because P$ lies on ellipse $\Rightarrow \frac{\alpha^{2}}{25} + \frac{\beta^{2}}{9} = 1$
$${\because PS + {PS}' = 2a \Rightarrow PS + {PS}' = 10 }{\therefore(PS)^{2} + \left( {PS}' \right)^{2} - {PS}'{PS}' = 37 }{\left( PS + {PS}' \right)^{2} - 3PS.{PS}' = 37 }{100 - 3PS.{PS}' = 37 }$$3PS.PS' $= 63 \Rightarrow$ PS.PS $= 21$
$\because PS\&{PS}'$ are $\left( 5 \pm \frac{4}{5} \cdot \alpha \right)$
∴ PS.PS $\ ' = 25 - \frac{16}{25}\alpha^{2} = 21$
$${\frac{16}{25}\alpha^{2} = 4 }{\alpha = \frac{5}{2} \Rightarrow \alpha^{2} = \frac{25}{4} }{\therefore\beta^{2} = \frac{27}{4} }{\therefore\alpha^{2} + \beta^{2} = \frac{52}{4} = 13}$$
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