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Question Number 131584 by mr W last updated on 06/Feb/21

Commented by mr W last updated on 06/Feb/21

find R=?

$${find}\:{R}=? \\ $$

Answered by ajfour last updated on 06/Feb/21

Point of contact  P(t,t^2 )  (dy/dx)=2t=tan θ  Rsin θ=t  t^2 −Rcos θ=R  t^2 =(1+(1/( (√(1+4t^2 )))))((t(√(1+4t^2 )))/(2t))  ⇒  2t^2 =(√(4t^2 +1))+1  (2t^2 −1)^2 =4t^2 +1  4t^4 −8t^2 =0  ⇒   t^2 =2  t=(√2)     R=(t/(2t))(√(1+4t^2 )) = (√((1/4)+2))  R=(3/2)  t^2 +Rcos θ=2+(1/2)  t+Rsin θ=(√2)+(√2)  Eq. of circles:  x^2 +(y−(5/2))^2 =(9/4)  (x−2(√2))^2 +(y−(3/2))^2 =(9/4)

$${Point}\:{of}\:{contact}\:\:{P}\left({t},{t}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{t}=\mathrm{tan}\:\theta \\ $$$${R}\mathrm{sin}\:\theta={t} \\ $$$${t}^{\mathrm{2}} −{R}\mathrm{cos}\:\theta={R} \\ $$$${t}^{\mathrm{2}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} }}\right)\frac{{t}\sqrt{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} }}{\mathrm{2}{t}} \\ $$$$\Rightarrow\:\:\mathrm{2}{t}^{\mathrm{2}} =\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1} \\ $$$$\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{4}{t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:{t}^{\mathrm{2}} =\mathrm{2} \\ $$$${t}=\sqrt{\mathrm{2}}\:\:\: \\ $$$${R}=\frac{{t}}{\mathrm{2}{t}}\sqrt{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} }\:=\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}} \\ $$$${R}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +{R}\mathrm{cos}\:\theta=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}+{R}\mathrm{sin}\:\theta=\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}} \\ $$$${Eq}.\:{of}\:{circles}: \\ $$$${x}^{\mathrm{2}} +\left({y}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\left({x}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$

Commented by ajfour last updated on 06/Feb/21

Commented by mr W last updated on 06/Feb/21

thanks sir!

$${thanks}\:{sir}! \\ $$

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