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
$$\mathrm{parabola}:\:{y}={ax}^{\mathrm{2}} \\ $$$$\mathrm{circle}:\:\left({x}−\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{{ap}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${r}=\sqrt{\left(\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{ap}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{{p}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{p}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\frac{{p}}{{a}^{\mathrm{2}} }{x}=\mathrm{0} \\ $$$$\mathrm{obviously}\:{x}_{\mathrm{1}} =\mathrm{0}\wedge{x}_{\mathrm{2}} ={p} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +{px}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{0}\wedge{x}_{\mathrm{3}} ={x}_{\mathrm{4}} \:\Rightarrow\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{2}}{{a}} \\ $$$$\Rightarrow\:{r}=\frac{\sqrt{\mathrm{5}}}{{a}} \\ $$
Commented by ajfour last updated on 09/Apr/20
$${Wise}\:{and}\:\mathcal{E}{legant}\:{way}\:{Sir}!\:{Thanks}. \\ $$
Commented by MJS last updated on 09/Apr/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}! \\ $$
Commented by jagoll last updated on 09/Apr/20
$$\mathrm{waww}...\mathrm{cooll} \\ $$