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Question-119395




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

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