Question Number 161296 by amin96 last updated on 15/Dec/21
Answered by mr W last updated on 16/Dec/21
$${A}\left({p}^{\mathrm{2}} ,{p}\right) \\ $$$${C}\left({q}^{\mathrm{2}} ,{q}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{1}}{\mathrm{2}{p}}=\mathrm{1} \\ $$$$\Rightarrow{p}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${BC}={q}−{p}=\frac{{AB}}{\mathrm{2}}=\frac{{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{q}+{p}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow{q}=\mathrm{2}−{p}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${BC}={q}−{p}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$
Commented by mr W last updated on 15/Dec/21
