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Question Number 144781 by imjagoll last updated on 29/Jun/21

Answered by liberty last updated on 29/Jun/21

$$\mid {\mathrm{PF}}_1\mid +\mid {\mathrm{PF}}_2\mid =2\mathrm{a}=6,\mathrm{by}\mathrm{definition}$$$$\mathrm{of}\mathrm{an}\mathrm{ellipse}.\mathrm{Since}\mid {\mathrm{PF}}_1\mid :\mid {\mathrm{PF}}_2\mid =2:1$$$$\mathrm{we}\mathrm{get}\{\begin{array}{l}\mid {\mathrm{PF}}_1\mid =4\\ \mid {\mathrm{PF}}_2\mid =2\end{array}$$$$\mathrm{notice}\mathrm{that}\mid {\mathrm{F}}_1{\mathrm{F}}_2\mid =2\mathrm{c}=2\sqrt{5}$$$$\mathrm{and}\mid {\mathrm{PF}}_1{\mid }^2+\mid {\mathrm{PF}}_2{\mid }^2=\mid {\mathrm{F}}_1{\mathrm{F}}_2{\mid }^2$$$$\mathrm{it}\mathrm{follows}\mathrm{that}△{\mathrm{PF}}_1{\mathrm{F}}_2\mathrm{is}\mathrm{a}\mathrm{right}$$$$\mathrm{triangle}\mathrm{so}\mathrm{the}\mathrm{area}\mathrm{of}△{\mathrm{PF}}_1{\mathrm{F}}_2=\frac{1}{2}\times 4\times 2=4.$$

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