Menu Close

Category: Geometry

Question-196829

Question Number 196829 by cortano12 last updated on 01/Sep/23 Answered by universe last updated on 01/Sep/23 $${by}\:{apollonius}\:{theorem} \\ $$$${BC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}\:} \:=\:\mathrm{2}\left({QC}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \right)\:\:\:…..\left(\mathrm{1}\right) \\ $$$${AC}^{\mathrm{2}}…

prove-that-the-curve-x-1-2-y-2-x-1-2-y-2-4-is-an-ellipse-and-find-its-semi-major-axis-and-semi-minor-axis-

Question Number 196723 by mr W last updated on 30/Aug/23 $${prove}\:{that}\:{the}\:{curve}\: \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4}\: \\ $$$${is}\:{an}\:{ellipse}\:{and}\:{find}\:{its}\:{semi} \\ $$$${major}\:{axis}\:{and}\:{semi}\:{minor}\:{axis}. \\ $$ Answered by…

Question-196275

Question Number 196275 by AROUNAMoussa last updated on 21/Aug/23 Answered by MM42 last updated on 21/Aug/23 $$\langle{ABD}=\mathrm{35}\Rightarrow{AD}={AB} \\ $$$$\langle{BCD}=\mathrm{55} \\ $$$$\frac{{AB}}{{AC}}=\frac{{sinx}}{{sin}\mathrm{130}}\:\:\:\&\:\:\frac{{AD}}{{AC}}=\frac{{sin}\left(\mathrm{55}−{x}\right)}{{sin}\mathrm{65}} \\ $$$$\Rightarrow{sinx}×{sin}\mathrm{65}=\mathrm{2}{sin}\mathrm{65}×{cos}\mathrm{65}×{sin}\left(\mathrm{55}−{x}\right) \\ $$$$\Rightarrow{sinx}=\mathrm{2}{cos}\mathrm{65}{sin}\left(\mathrm{55}−{x}\right)={sin}\left(\mathrm{55}−{x}+\mathrm{65}\right)+{sin}\left(\mathrm{55}−{x}−\mathrm{65}\right)…