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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

$$sayy={ax}^2$$$$P(p,{ap}^2)$$$$Q(-q,{aq}^2)$$$$AP\mathrm{\perp }AQ$$$$\Rightarrow \frac{{ap}^2}{p}\times \frac{{aq}^2}{-q}=-1$$$$\Rightarrow pq=\frac{1}{a^2}$$$$eqn.ofPQ:$$$$\frac{y-{aq}^2}{{ap}^2-{aq}^2}=\frac{x+q}{p+q}$$$$intersectionpointR:$$$$\frac{y_R-{aq}^2}{{ap}^2-{aq}^2}=\frac{0+q}{p+q}$$$$\frac{y_R}{a}-q^2=q(p-q)$$$$\frac{y_R}{a}=pq=\frac{1}{a^2}$$$$\Rightarrow y_R=\frac{1}{a}=constant$$$$i.e.R(0,\frac{1}{a})isafixedpoint.$$$$AR=y_R=\frac{1}{a}$$

Commented by peter frank last updated on 24/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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