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

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

$$P(p,\frac{c^2}{p})$$$$Q(q,\frac{c^2}{q})$$$$PQ=\sqrt{{(p-q)}^2+{(\frac{c^2}{p}-\frac{c^2}{q})}^2}=k$$$$\Rightarrow {(p-q)}^2(1+\frac{c^4}{p^2{q}^2})=k^2...\left(i\right)$$$$letR(s,t)$$$$s=\frac{1}{2}(p+q)$$$$t=\frac{1}{2}(\frac{c^2}{p}+\frac{c^2}{q})=\frac{c^2(p+q)}{2pq}$$$$\Rightarrow p+q=2s$$$$\Rightarrow t=\frac{c^{2}s}{pq}$$$$\Rightarrow pq=\frac{c^{2}s}{t}$$$$\Rightarrow {(p-q)}^2={(p+q)}^2-4pq=4s^2-\frac{4c^{2}s}{t}=4s(s-\frac{c^2}{t})$$$$puttheminto\left(i\right):$$$$\Rightarrow 4s(s-\frac{c^2}{t})(1+\frac{t^2}{s^2})=k^2$$$$\Rightarrow 4(st-c^2)(s^2+t^2)=k^{2}st$$$$or$$$$\Rightarrow 4(xy-c^2)(x^2+y^2)=k^{2}xy$$

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

Commented by peter frank last updated on 16/Dec/18

$$\mathrm{sir}\mathrm{please}\mathrm{check}\mathrm{my}\mathrm{solution}\mathrm{QN}50328$$

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