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Question-155772




Question Number 155772 by daus last updated on 04/Oct/21
Answered by mr W last updated on 04/Oct/21
Commented by mr W last updated on 04/Oct/21
T((b/( (√2))),(b/( (√2))))  S(a,0)  P(x,y)  y=((0+(b/( (√2))))/2)=(b/(2(√2))) ⇒b=2(√2)y  x=((a+(b/( (√2))))/2)=(a/2)+y ⇒a=2(x−y)  let ST=l=2  l^2 =a^2 +b^2 −(√2)ab  l^2 =8y^2 +4(x−y)^2 −8y(x−y)  ⇒x^2 +5y^2 −4xy=((l/2))^2   this is an ellipse.
$${T}\left(\frac{{b}}{\:\sqrt{\mathrm{2}}},\frac{{b}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${S}\left({a},\mathrm{0}\right) \\ $$$${P}\left({x},{y}\right) \\ $$$${y}=\frac{\mathrm{0}+\frac{{b}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}}=\frac{{b}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow{b}=\mathrm{2}\sqrt{\mathrm{2}}{y} \\ $$$${x}=\frac{{a}+\frac{{b}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}}=\frac{{a}}{\mathrm{2}}+{y}\:\Rightarrow{a}=\mathrm{2}\left({x}−{y}\right) \\ $$$${let}\:{ST}={l}=\mathrm{2} \\ $$$${l}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\sqrt{\mathrm{2}}{ab} \\ $$$${l}^{\mathrm{2}} =\mathrm{8}{y}^{\mathrm{2}} +\mathrm{4}\left({x}−{y}\right)^{\mathrm{2}} −\mathrm{8}{y}\left({x}−{y}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{5}{y}^{\mathrm{2}} −\mathrm{4}{xy}=\left(\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${this}\:{is}\:{an}\:{ellipse}. \\ $$
Commented by mr W last updated on 04/Oct/21
Commented by daus last updated on 05/Oct/21
thanks
$${thanks}\: \\ $$

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