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Question Number 215520 by CrispyXYZ last updated on 09/Jan/25
Prove that { ((x = ((7 cos t − 2)/(2 − cos t)))),((y = ((4(√3) sin t)/(2 − cos t)))) :} is a circle.
$$\mathrm{Prove}\:\mathrm{that}\:\begin{cases}{{x}\:=\:\frac{\mathrm{7}\:\mathrm{cos}\:{t}\:−\:\mathrm{2}}{\mathrm{2}\:−\:\mathrm{cos}\:{t}}}\\{{y}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{t}}{\mathrm{2}\:−\:\mathrm{cos}\:{t}}}\end{cases}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}. \\ $$
Answered by alephnull last updated on 09/Jan/25
= x(2−cos(t))=7cos(t)−2, y(2−cos (t))=4(√3)sin (t) Rearrange 2x−xcos (t)=7cos (t)−2 2y−ycos (t)=4(√3)sin (t) ↓ xcos (t)=(7cos (t)−2)−2x ycos (t)=4(√3)sin (t)−2y recall pythagoras indentity cos^2 t+sin^2 t=1 cos(t)=((2x+2)/(x−7)), sin(t)=(((2−x)y)/(4(√3))) Substitute (((2x+2)/(x−7)))^2 +((((2−x)y)/(4(√3))))^2 =1 simplify (((2x+2)^2 )/((x−7)^2 ))+(((2−x)^2 y^2 )/(48))=1 =48(2x+2)^2 +(2−x)^2 y^2 (x−7)^2 =48(x−7)^2 answer resembles circle thing (x−h)^2 +(y−k)^2 =r so it is
$$=\:{x}\left(\mathrm{2}−\mathrm{cos}\left({t}\right)\right)=\mathrm{7cos}\left({t}\right)−\mathrm{2},\:{y}\left(\mathrm{2}−\mathrm{cos}\:\left({t}\right)\right)=\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}\:\left({t}\right) \\ $$$$\mathrm{Rearrange} \\ $$$$\mathrm{2}{x}−{x}\mathrm{cos}\:\left({t}\right)=\mathrm{7cos}\:\left({t}\right)−\mathrm{2} \\ $$$$\mathrm{2}{y}−{y}\mathrm{cos}\:\left({t}\right)=\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}\:\left({t}\right) \\ $$$$\downarrow \\ $$$${x}\mathrm{cos}\:\left({t}\right)=\left(\mathrm{7cos}\:\left({t}\right)−\mathrm{2}\right)−\mathrm{2}{x} \\ $$$${y}\mathrm{cos}\:\left({t}\right)=\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}\:\left({t}\right)−\mathrm{2}{y} \\ $$$$ \\ $$$$\mathrm{recall}\:\mathrm{pythagoras}\:\mathrm{indentity} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {t}+\mathrm{sin}\:^{\mathrm{2}} {t}=\mathrm{1} \\ $$$$ \\ $$$${cos}\left({t}\right)=\frac{\mathrm{2}{x}+\mathrm{2}}{{x}−\mathrm{7}},\:{sin}\left({t}\right)=\frac{\left(\mathrm{2}−{x}\right){y}}{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{S}{ubstitute} \\ $$$$\left(\frac{\mathrm{2}{x}+\mathrm{2}}{{x}−\mathrm{7}}\right)^{\mathrm{2}} +\left(\frac{\left(\mathrm{2}−{x}\right){y}}{\mathrm{4}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{simplify} \\ $$$$\frac{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }{\left({x}−\mathrm{7}\right)^{\mathrm{2}} }+\frac{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{48}}=\mathrm{1} \\ $$$$=\mathrm{48}\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−{x}\right)^{\mathrm{2}} {y}^{\mathrm{2}} \left({x}−\mathrm{7}\right)^{\mathrm{2}} =\mathrm{48}\left({x}−\mathrm{7}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{answer}\:\mathrm{resembles}\:\mathrm{circle}\:\mathrm{thing} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{is} \\ $$$$ \\ $$

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