Menu Close

Question-135952




Question Number 135952 by Dwaipayan Shikari last updated on 17/Mar/21
Commented by Dwaipayan Shikari last updated on 17/Mar/21
It isArchimedes′s Nothing Grinder. We can move this back  and forth and sideways .Prove that the tip on various  positions on that hand of the wodden grinder creates   various shapes of ellipses
$${It}\:{isArchimedes}'{s}\:{Nothing}\:{Grinder}.\:{We}\:{can}\:{move}\:{this}\:{back} \\ $$$${and}\:{forth}\:{and}\:{sideways}\:.{Prove}\:{that}\:{the}\:{tip}\:{on}\:{various} \\ $$$${positions}\:{on}\:{that}\:{hand}\:{of}\:{the}\:{wodden}\:{grinder}\:{creates}\: \\ $$$${various}\:{shapes}\:{of}\:{ellipses} \\ $$
Commented by Dwaipayan Shikari last updated on 17/Mar/21
Take a look at this animation https://en.wikipedia.org/wiki/File:Trammel_of_Archimedes_Large.gif
Answered by mr W last updated on 17/Mar/21
Commented by mr W last updated on 17/Mar/21
say  B(t,0)  A(0,s)  s^2 +t^2 =p^2   x_C =((q+p)/p)t ⇒t=(p/(p+q))x_C   y_C =−(q/p)s ⇒s=−(p/q)y_C   ((p/(p+q))x_C )^2 +(−(p/q)y_C )^2 =p^2   (x_C ^2 /((p+q)^2 ))+(y_C ^2 /q^2 )=1  or  (x^2 /((p+q)^2 ))+(y^2 /q^2 )=1  i.e. the locus of C is an ellipse with  semiaxses:  a=p+q  b=q
$${say} \\ $$$${B}\left({t},\mathrm{0}\right) \\ $$$${A}\left(\mathrm{0},{s}\right) \\ $$$${s}^{\mathrm{2}} +{t}^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${x}_{{C}} =\frac{{q}+{p}}{{p}}{t}\:\Rightarrow{t}=\frac{{p}}{{p}+{q}}{x}_{{C}} \\ $$$${y}_{{C}} =−\frac{{q}}{{p}}{s}\:\Rightarrow{s}=−\frac{{p}}{{q}}{y}_{{C}} \\ $$$$\left(\frac{{p}}{{p}+{q}}{x}_{{C}} \right)^{\mathrm{2}} +\left(−\frac{{p}}{{q}}{y}_{{C}} \right)^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$$\frac{{x}_{{C}} ^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} }+\frac{{y}_{{C}} ^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\mathrm{1} \\ $$$${or} \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{locus}\:{of}\:{C}\:{is}\:{an}\:{ellipse}\:{with} \\ $$$${semiaxses}: \\ $$$${a}={p}+{q} \\ $$$${b}={q} \\ $$
Commented by Dwaipayan Shikari last updated on 17/Mar/21
Thanks sir! i didn′t think that the proof is simple
$${Thanks}\:{sir}!\:{i}\:{didn}'{t}\:{think}\:{that}\:{the}\:{proof}\:{is}\:{simple} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *