Question Number 190019 by ajfour last updated on 26/Mar/23
Commented by ajfour last updated on 26/Mar/23
$${Given}\:{p}\:{and}\:{q}.\:{Find}\:{r}.\: \\ $$$${Parabola}\:{is}\:{y}={x}^{\mathrm{2}} . \\ $$
Commented by mr W last updated on 26/Mar/23
$${you}\:{have}\:{given} \\ $$$${r}=\frac{\sqrt{{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 26/Mar/23
$${oops}!,\:{forget}\:{q}\:{being}\:{givwn}. \\ $$
Answered by ajfour last updated on 26/Mar/23
$${A}\left(−{s},{s}^{\mathrm{2}} \right)\:\:\:\:;\:\mathrm{tan}\:\alpha=\mathrm{2}{s} \\ $$$${B}\left({t},{t}^{\mathrm{2}} \right)\:\:;\:\mathrm{tan}\:\beta=\mathrm{2}{t} \\ $$$$−{s}+{r}\mathrm{sin}\:\alpha={p}\mathrm{cos}\:\theta \\ $$$${s}^{\mathrm{2}} +{r}\mathrm{cos}\:\alpha={p}\mathrm{sin}\:\theta \\ $$$$−{p}\mathrm{cos}\:\theta+\mathrm{2}{r}\mathrm{sin}\:\theta+{r}\mathrm{sin}\:\beta={t} \\ $$$${p}\mathrm{sin}\:\theta+\mathrm{2}{r}\mathrm{cos}\:\theta−{r}\mathrm{cos}\:\beta={t}^{\mathrm{2}} \\ $$$$.. \\ $$