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Question-74817




Question Number 74817 by peter frank last updated on 01/Dec/19
Answered by mr W last updated on 03/Dec/19
parallel chords:  y=mx+k  x(mx+k)=c^2   mx^2 +kx−c^2 =0  x_1 +x_2 =−(k/m)  x_1 x_2 =−(c^2 /m)  middle point of chord: M(p,q)  p=((x_1 +x_2 )/2)=−(k/(2m))  q=((y_1 +y_2 )/2)=(c^2 /2)((1/x_1 )+(1/x_2 ))=((c^2 (x_1 +x_2 ))/(2x_1 x_2 ))=(k/2)  ⇒q=−mp  ⇒y=−mx ← eqn. of locus
$${parallel}\:{chords}: \\ $$$${y}={mx}+{k} \\ $$$${x}\left({mx}+{k}\right)={c}^{\mathrm{2}} \\ $$$${mx}^{\mathrm{2}} +{kx}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\frac{{k}}{{m}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =−\frac{{c}^{\mathrm{2}} }{{m}} \\ $$$${middle}\:{point}\:{of}\:{chord}:\:{M}\left({p},{q}\right) \\ $$$${p}=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}=−\frac{{k}}{\mathrm{2}{m}} \\ $$$${q}=\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} }{\mathrm{2}}=\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}_{\mathrm{1}} }+\frac{\mathrm{1}}{{x}_{\mathrm{2}} }\right)=\frac{{c}^{\mathrm{2}} \left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)}{\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} }=\frac{{k}}{\mathrm{2}} \\ $$$$\Rightarrow{q}=−{mp} \\ $$$$\Rightarrow{y}=−{mx}\:\leftarrow\:{eqn}.\:{of}\:{locus} \\ $$
Commented by peter frank last updated on 06/Dec/19
thank you
$${thank}\:{you}\: \\ $$

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