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Question-185151




Question Number 185151 by mathlove last updated on 17/Jan/23
Answered by mr W last updated on 20/Jan/23
say the touching point with the  parabola is (−p,(p^2 /2)).  the center of circle is (−R,h).  tan θ=p  −R=−p+R sin θ=−p+((Rp)/( (√(p^2 +1))))  ⇒R=(p/(1+(p/( (√(p^2 +1))))))  h=(p^2 /2)+R cos θ=(p^2 /2)+(R/( (√(p^2 +1))))  ⇒h=(p^2 /2)+(p/(p+(√(p^2 +1))))  eqn. of tangent line:  y=2−(x−4)(1/( (√3)))  x+(√3)y−2(√3)−4=0  (((−R+(√3)h−2(√3)−4))/( (√(1^2 +((√3))^2 ))))=−R  (√3)h+R−2(√3)−4=0  (√3)((p^2 /2)+(p/(p+(√(p^2 +1)))))+((p/(1+(p/( (√(p^2 +1)))))))−2(√3)−4=0  ⇒p≈2.4837  ⇒R≈1.2885
$${say}\:{the}\:{touching}\:{point}\:{with}\:{the} \\ $$$${parabola}\:{is}\:\left(−{p},\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right). \\ $$$${the}\:{center}\:{of}\:{circle}\:{is}\:\left(−{R},{h}\right). \\ $$$$\mathrm{tan}\:\theta={p} \\ $$$$−{R}=−{p}+{R}\:\mathrm{sin}\:\theta=−{p}+\frac{{Rp}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{R}=\frac{{p}}{\mathrm{1}+\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}} \\ $$$${h}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+{R}\:\mathrm{cos}\:\theta=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{h}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}}{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${eqn}.\:{of}\:{tangent}\:{line}: \\ $$$${y}=\mathrm{2}−\left({x}−\mathrm{4}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${x}+\sqrt{\mathrm{3}}{y}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\frac{\left(−{R}+\sqrt{\mathrm{3}}{h}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}\right)}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}=−{R} \\ $$$$\sqrt{\mathrm{3}}{h}+{R}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\sqrt{\mathrm{3}}\left(\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}}{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}\right)+\left(\frac{{p}}{\mathrm{1}+\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}}\right)−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{2}.\mathrm{4837} \\ $$$$\Rightarrow{R}\approx\mathrm{1}.\mathrm{2885} \\ $$
Commented by mr W last updated on 20/Jan/23

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