Question-50248

Question Number 50248 by peter frank last updated on 15/Dec/18

Answered by mr W last updated on 15/Dec/18

P(p,(c^2 /p)) Q(q,(c^2 /q)) PQ=(√((p−q)^2 +((c^2 /p)−(c^2 /q))^2 ))=k ⇒(p−q)^2 (1+(c^4 /(p^2 q^2 )))=k^2 ...(i) let R(s,t) s=(1/2)(p+q) t=(1/2)((c^2 /p)+(c^2 /q))=((c^2 (p+q))/(2pq)) ⇒p+q=2s ⇒t=((c^2 s)/(pq)) ⇒pq=((c^2 s)/t) ⇒(p−q)^2 =(p+q)^2 −4pq=4s^2 −((4c^2 s)/t)=4s(s−(c^2 /t)) put them into (i): ⇒4s(s−(c^2 /t))(1+(t^2 /s^2 ))=k^2 ⇒4(st−c^2 )(s^2 +t^2 )=k^2 st or ⇒4(xy−c^2 )(x^2 +y^2 )=k^2 xy

$${P}\left({p},\frac{{c}^{\mathrm{2}} }{{p}}\right) \\ $$$${Q}\left({q},\frac{{c}^{\mathrm{2}} }{{q}}\right) \\ $$$${PQ}=\sqrt{\left({p}−{q}\right)^{\mathrm{2}} +\left(\frac{{c}^{\mathrm{2}} }{{p}}−\frac{{c}^{\mathrm{2}} }{{q}}\right)^{\mathrm{2}} }={k} \\ $$$$\Rightarrow\left({p}−{q}\right)^{\mathrm{2}} \left(\mathrm{1}+\frac{{c}^{\mathrm{4}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} }\right)={k}^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$${let}\:{R}\left({s},{t}\right) \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left({p}+{q}\right) \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{c}^{\mathrm{2}} }{{p}}+\frac{{c}^{\mathrm{2}} }{{q}}\right)=\frac{{c}^{\mathrm{2}} \left({p}+{q}\right)}{\mathrm{2}{pq}} \\ $$$$\Rightarrow{p}+{q}=\mathrm{2}{s} \\ $$$$\Rightarrow{t}=\frac{{c}^{\mathrm{2}} {s}}{{pq}} \\ $$$$\Rightarrow{pq}=\frac{{c}^{\mathrm{2}} {s}}{{t}} \\ $$$$\Rightarrow\left({p}−{q}\right)^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{4}{pq}=\mathrm{4}{s}^{\mathrm{2}} −\frac{\mathrm{4}{c}^{\mathrm{2}} {s}}{{t}}=\mathrm{4}{s}\left({s}−\frac{{c}^{\mathrm{2}} }{{t}}\right) \\ $$$${put}\:{them}\:{into}\:\left({i}\right): \\ $$$$\Rightarrow\mathrm{4}{s}\left({s}−\frac{{c}^{\mathrm{2}} }{{t}}\right)\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\right)={k}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}\left({st}−{c}^{\mathrm{2}} \right)\left({s}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)={k}^{\mathrm{2}} {st} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{4}\left({xy}−{c}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)={k}^{\mathrm{2}} {xy} \\ $$

Commented by mr W last updated on 15/Dec/18