Question-88240

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Question Number 88240 by ajfour last updated on 09/Apr/20

Commented by ajfour last updated on 09/Apr/20

$${Find}\:{radius}\:{of}\:{semicircle}\:{in} \\ $$$${terms}\:{of}\:{a}. \\ $$

Commented by ajfour last updated on 09/Apr/20

$${mjS}\:{Sir},\:{wd}\:{u}\:{like}\:{solving}\:{this}. \\ $$

Commented by MJS last updated on 09/Apr/20

$$\mathrm{let}\:\mathrm{me}\:\mathrm{try}… \\ $$

Answered by MJS last updated on 09/Apr/20

parabola: y=ax^2 circle: (x−(p/2))^2 +(y−((ap^2 )/2))^2 =r^2 r=(√(((p/2))^2 +(((ap^2 )/2))^2 ))=(p/2)(√(a^2 p^2 +1)) ⇒ x^4 +((1/a^2 )−p^2 )x^2 −(p/a^2 )x=0 obviously x_1 =0∧x_2 =p ⇒ x^2 +px+(1/a^2 )=0∧x_3 =x_4 ⇒ D=0 ⇒ p=(2/a) ⇒ r=((√5)/a)

$$\mathrm{parabola}:\:{y}={ax}^{\mathrm{2}} \\ $$$$\mathrm{circle}:\:\left({x}−\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{{ap}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${r}=\sqrt{\left(\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{ap}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{{p}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{p}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\frac{{p}}{{a}^{\mathrm{2}} }{x}=\mathrm{0} \\ $$$$\mathrm{obviously}\:{x}_{\mathrm{1}} =\mathrm{0}\wedge{x}_{\mathrm{2}} ={p} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +{px}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{0}\wedge{x}_{\mathrm{3}} ={x}_{\mathrm{4}} \:\Rightarrow\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{2}}{{a}} \\ $$$$\Rightarrow\:{r}=\frac{\sqrt{\mathrm{5}}}{{a}} \\ $$

Commented by ajfour last updated on 09/Apr/20