Question Number 47454 by peter frank last updated on 10/Nov/18 | ||
$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{middle}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{normal}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{parabola}\:\mathrm{y}^{\mathrm{2}} =\mathrm{4ax}\:\mathrm{is} \\ $$$$\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2a}}+\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{y}^{\mathrm{2}} }=\mathrm{x}−\mathrm{2a} \\ $$ | ||
Answered by ajfour last updated on 10/Nov/18 | ||
$${M}\left({h},{k}\right) \\ $$$${y}=\mathrm{2}{at}\:\:;\:{x}={at}^{\mathrm{2}} \\ $$$${M}\:{is}\:{midpoint}\:{of}\:{chord},\:{so} \\ $$$$\Rightarrow\:\:{k}−\mathrm{2}{at}_{\mathrm{1}} =\:−{t}_{\mathrm{1}} \left({h}−{at}_{\mathrm{1}} ^{\mathrm{2}} \right)\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$$\:\:\:\mathrm{2}{a}\left({t}_{\mathrm{2}} +{t}_{\mathrm{1}} \right)=\mathrm{2}{k}\:\:\:\:\:\:...\left({ii}\right) \\ $$$$\:\:\:\:{a}\left({t}_{\mathrm{2}} ^{\mathrm{2}} +{t}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mathrm{2}{h}\:\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2}{a}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}{{a}\left({t}_{\mathrm{2}} ^{\mathrm{2}} −{t}_{\mathrm{1}} ^{\mathrm{2}} \right)}=\:−{t}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:{t}_{\mathrm{1}} \left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)=\:−\mathrm{2}\:\:\:\:\:\:...\left({iv}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({iv}\right) \\ $$$$\:\:\:\:\:{t}_{\mathrm{1}} \left(\frac{{k}}{{a}}\right)=−\mathrm{2} \\ $$$$\Rightarrow\:\:\:{t}_{\mathrm{1}} \:=\:\frac{−\mathrm{2}{a}}{{k}} \\ $$$${using}\:{this}\:{in}\:\left({i}\right) \\ $$$$\:\:{k}−\mathrm{2}{a}\left(\frac{−\mathrm{2}{a}}{{k}}\right)=\:\left(\frac{\mathrm{2}{a}}{{k}}\right)\left[{h}−{a}\left(\frac{\mathrm{4}{a}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)\right] \\ $$$${Now}\:\left({h},{k}\right)\rightarrow\left({x},{y}\right) \\ $$$$\left({y}+\frac{\mathrm{4}{a}^{\mathrm{2}} }{{y}}\right)\left(\frac{{y}}{\mathrm{2}{a}}\right)=\:{x}−\frac{\mathrm{4}{a}^{\mathrm{3}} }{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}{a}}+\mathrm{2}{a}\:=\:{x}−\frac{\mathrm{4}{a}^{\mathrm{3}} }{{y}^{\mathrm{2}} } \\ $$$${or}\:\:\:\:\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}{a}}\:+\:\frac{\mathrm{4}{a}^{\mathrm{3}} }{{y}^{\mathrm{2}} }\:=\:{x}−\mathrm{2}{a}\:. \\ $$ | ||