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Question-226855




Question Number 226855 by Kassista last updated on 17/Dec/25
Answered by mehdee7396 last updated on 17/Dec/25
x^2 +4x^2 −4x+4=4⇒x=0,(8/5)  ⇒P(0,,2) &  Q((8/5),−(6/5))⇒m_(PQ) =−(3/4)  ⇒tanα=(3/4)⋍36^0 ⇒∠POQ⋍126^0
$${x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{4}\Rightarrow{x}=\mathrm{0},\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\Rightarrow{P}\left(\mathrm{0},,\mathrm{2}\right)\:\&\:\:{Q}\left(\frac{\mathrm{8}}{\mathrm{5}},−\frac{\mathrm{6}}{\mathrm{5}}\right)\Rightarrow{m}_{{PQ}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}}\backsimeq\mathrm{36}^{\mathrm{0}} \Rightarrow\angle{POQ}\backsimeq\mathrm{126}^{\mathrm{0}} \\ $$

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