Question Number 212714 by Mingma last updated on 21/Oct/24
Answered by mr W last updated on 22/Oct/24
Commented by mr W last updated on 22/Oct/24
$${say}\:{P}\left({p},\:{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$${r}=\frac{{p}}{\mathrm{sin}\:\theta}=\frac{\sqrt{\mathrm{15}}}{\mathrm{2}×\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}}=\mathrm{2}\:\checkmark \\ $$
Commented by Mingma last updated on 22/Oct/24
perfect sir
Answered by ajfour last updated on 22/Oct/24
$${q}={p}^{\mathrm{2}} \\ $$$${equation}\:{of}\:{common}\:{Normal}: \\ $$$${y}−{q}=\left(−\frac{\mathrm{1}}{\mathrm{2}{p}}\right)\left({x}−{p}\right) \\ $$$${set}\:\:{x}_{\mathrm{0}} =\mathrm{0}\:\:\Rightarrow\:\:{y}_{\mathrm{0}} −{q}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}=\mathrm{4}\left({y}_{\mathrm{0}} −{q}\right)=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2} \\ $$