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Question-144781




Question Number 144781 by imjagoll last updated on 29/Jun/21
Answered by liberty last updated on 29/Jun/21
 ∣PF_1 ∣+∣PF_2 ∣=2a =6 , by definition  of an ellipse. Since ∣PF_1 ∣:∣PF_2 ∣=2:1  we get  { ((∣PF_1 ∣=4)),((∣PF_2 ∣=2)) :}  notice that ∣F_1 F_2 ∣=2c = 2(√5)  and ∣PF_1 ∣^2 +∣PF_2 ∣^2 =∣F_1 F_2 ∣^2   it follows that △PF_1 F_2  is a right  triangle so the area of △PF_1 F_2 =(1/2)×4×2=4.
$$\:\mid\mathrm{PF}_{\mathrm{1}} \mid+\mid\mathrm{PF}_{\mathrm{2}} \mid=\mathrm{2a}\:=\mathrm{6}\:,\:\mathrm{by}\:\mathrm{definition} \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}.\:\mathrm{Since}\:\mid\mathrm{PF}_{\mathrm{1}} \mid:\mid\mathrm{PF}_{\mathrm{2}} \mid=\mathrm{2}:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mid\mathrm{PF}_{\mathrm{1}} \mid=\mathrm{4}}\\{\mid\mathrm{PF}_{\mathrm{2}} \mid=\mathrm{2}}\end{cases} \\ $$$$\mathrm{notice}\:\mathrm{that}\:\mid\mathrm{F}_{\mathrm{1}} \mathrm{F}_{\mathrm{2}} \mid=\mathrm{2c}\:=\:\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{and}\:\mid\mathrm{PF}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid\mathrm{PF}_{\mathrm{2}} \mid^{\mathrm{2}} =\mid\mathrm{F}_{\mathrm{1}} \mathrm{F}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\bigtriangleup\mathrm{PF}_{\mathrm{1}} \mathrm{F}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{right} \\ $$$$\mathrm{triangle}\:\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\bigtriangleup\mathrm{PF}_{\mathrm{1}} \mathrm{F}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\mathrm{2}=\mathrm{4}. \\ $$

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