Question Number 119395 by peter frank last updated on 24/Oct/20
Answered by mr W last updated on 24/Oct/20
Commented by mr W last updated on 24/Oct/20
$${say}\:{y}={ax}^{\mathrm{2}} \\ $$$${P}\left({p},{ap}^{\mathrm{2}} \right) \\ $$$${Q}\left(−{q},{aq}^{\mathrm{2}} \right) \\ $$$${AP}\bot{AQ} \\ $$$$\Rightarrow\frac{{ap}^{\mathrm{2}} }{{p}}×\frac{{aq}^{\mathrm{2}} }{−{q}}=−\mathrm{1} \\ $$$$\Rightarrow{pq}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$${eqn}.\:{of}\:{PQ}: \\ $$$$\frac{{y}−{aq}^{\mathrm{2}} }{{ap}^{\mathrm{2}} −{aq}^{\mathrm{2}} }=\frac{{x}+{q}}{{p}+{q}} \\ $$$${intersection}\:{point}\:{R}: \\ $$$$\frac{{y}_{{R}} −{aq}^{\mathrm{2}} }{{ap}^{\mathrm{2}} −{aq}^{\mathrm{2}} }=\frac{\mathrm{0}+{q}}{{p}+{q}} \\ $$$$\frac{{y}_{{R}} }{{a}}−{q}^{\mathrm{2}} ={q}\left({p}−{q}\right) \\ $$$$\frac{{y}_{{R}} }{{a}}={pq}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{y}_{{R}} =\frac{\mathrm{1}}{{a}}={constant} \\ $$$${i}.{e}.\:{R}\left(\mathrm{0},\frac{\mathrm{1}}{{a}}\right)\:{is}\:{a}\:{fixed}\:{point}. \\ $$$${AR}={y}_{{R}} =\frac{\mathrm{1}}{{a}} \\ $$
Commented by peter frank last updated on 24/Oct/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$