Let S and S^' be the foci of the ellipse x^225+y^29=1 and P( , ) be a point on the ellipse…

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 $(SP{)}^{2}+{({S}^{'}P)}^{2}-SP\cdot {S}^{'}P=37$, then ${\alpha }^{2}+{\beta }^{2}$ is equal to :

Accepted Answer

$\begin{gathered} \because P\Rightarrow \frac{{\alpha }^{2}}{25}+\frac{{\beta }^{2}}{9}=1 \ \because PS+P{S}^{'}=2a\Rightarrow PS+P{S}^{'}=10 \ \because (PS{)}^{2}+{(P{S}^{'})}^{2}-PS.P{S}^{'}=37 \ (PS+P{S}^{'})-3PS.P{S}^{'}=37 \ 100-3PS.P{S}^{'}=37 \ 3PS.P{S}^{'}=63\Rightarrow PS.P{S}^{'}=21 \ \because PSP{S}^{'}(5\pm \frac{4}{5}.\alpha ) \ 	herefore PS.P{S}^{'}=25-\frac{16}{25}{\alpha }^{2}=21 \ \frac{16}{25}{\alpha }^{2}=4 \ \alpha =\frac{5}{2}\Rightarrow {\alpha }^{2}=\frac{25}{4} \ 	herefore {\beta }^{2}=\frac{27}{4} \ 	herefore {\alpha }^{2}+{\beta }^{2}=\frac{52}{4}=13 \end{gathered}$

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