Question Number 27818 by ajfour last updated on 15/Jan/18
Commented by ajfour last updated on 15/Jan/18
$${If}\:{y}={Ax}^{\mathrm{2}} \:{is}\:{reflected}\:{in}\:{the}\:{line} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:,\:{find}\:{equation}\:{of}\:{its} \\ $$$${reflection}\:{f}\left({x},{y}\right)=\mathrm{0}\:. \\ $$
Commented by mrW2 last updated on 15/Jan/18
$${let}'{s}\:{say}\:{point}\:{A}\left({p},{q}\right)\:{is}\:{reflected}\:{in} \\ $$$${line}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{and}\:{its}\:{mirrored}\:{image} \\ $$$${is}\:{point}\:{C}\left({u},{v}\right). \\ $$$${let}'{s}\:{say}\:{B}\:{is}\:{a}\:{point}\:{on}\:{the}\:{line}\:{with} \\ $$$${AB}\bot{to}\:{the}\:{line}. \\ $$$$ \\ $$$${Eqn}.\:{of}\:{AB}\:{is} \\ $$$$\left({x},{y}\right)=\left({p},{r}\right)+\lambda\left(\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}}\right) \\ $$$${or} \\ $$$${x}={p}+\frac{\lambda}{{a}} \\ $$$${y}={q}+\frac{\lambda}{{b}} \\ $$$${point}\:{B}\:{is}\:{on}\:{the}\:{line}: \\ $$$$\frac{\mathrm{1}}{{a}}\left({p}+\frac{\lambda_{{B}} }{{a}}\right)+\frac{\mathrm{1}}{{b}}\left({q}+\frac{\lambda_{{B}} }{{b}}\right)=\mathrm{1} \\ $$$$\lambda_{{B}} \left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)=\mathrm{1}−\left(\frac{{p}}{{a}}+\frac{{q}}{{b}}\right) \\ $$$$\lambda_{{B}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}\right)={ab}−\left({aq}+{bp}\right) \\ $$$$\Rightarrow\lambda_{{B}} =\frac{{ab}\left({ab}−{aq}−{bp}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${since}\:{C}\:{is}\:{mirrored}\:{image}\:{of}\:{A}, \\ $$$$\lambda_{{C}} =\mathrm{2}\lambda_{{B}} =\frac{\mathrm{2}{ab}\left({ab}−{aq}−{bp}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}={p}+\frac{\lambda_{{C}} }{{a}}={p}+\frac{\mathrm{2}{b}\left({ab}−{aq}−{bp}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{p}−\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{q}+\frac{\mathrm{2}{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={cp}+{dq}−{db} \\ $$$$\Rightarrow{v}={q}+\frac{\lambda_{{C}} }{{b}}={q}+\frac{\mathrm{2}{a}\left({ab}−{aq}−{bp}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=−\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{p}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{q}+\frac{\mathrm{2}{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={dp}−{cq}−{da} \\ $$$${with}\:{c}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:{d}=−\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${now}\:{we}\:{have}\:\left({u},{v}\right)\:{on}\:{the}\:{curve}\:{y}={Ax}^{\mathrm{2}} \\ $$$$\Rightarrow{v}={Au}^{\mathrm{2}} \\ $$$${dp}−{cq}−{da}={A}\left({cp}+{dq}−{db}\right)^{\mathrm{2}} ={A}\left({c}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{2}{cdpq}+{d}^{\mathrm{2}} {q}^{\mathrm{2}} +{d}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{dbcp}−\mathrm{2}{d}^{\mathrm{2}} {bq}\right) \\ $$$${Ac}^{\mathrm{2}} {p}^{\mathrm{2}} +{Ad}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{Acdpq}+{Ad}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{Adbcp}−\mathrm{2}{Ad}^{\mathrm{2}} {bq}−{dp}+{cq}+{da}=\mathrm{0} \\ $$$${Ac}^{\mathrm{2}} {p}^{\mathrm{2}} +{Ad}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{Acdpq}−{d}\left(\mathrm{1}+\mathrm{2}{Abc}\right){p}−\left(\mathrm{2}{Abd}^{\mathrm{2}} −{c}\right){q}+{d}\left({a}+{Adb}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${or} \\ $$$${Ac}^{\mathrm{2}} {x}^{\mathrm{2}} +{Ad}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{Acdxy}−{d}\left(\mathrm{1}+\mathrm{2}{Abc}\right){x}−\left(\mathrm{2}{Abd}^{\mathrm{2}} −{c}\right){y}+{d}\left({a}+{Adb}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\frac{{A}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{x}^{\mathrm{2}} +\frac{\mathrm{4}{Aa}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{y}^{\mathrm{2}} −\frac{\mathrm{4}{Aab}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{xy}+\mathrm{2}{ab}\left(\mathrm{1}+\frac{\mathrm{2}{Ab}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right){x}−\left(\frac{\mathrm{8}{Aa}^{\mathrm{2}} {b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){y}−\mathrm{2}{ab}\left({a}−\frac{\mathrm{2}{Aab}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left({x},{y}\right)={A}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{Aa}^{\mathrm{2}} {b}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{4}{Aab}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){xy}+\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{Aa}^{\mathrm{2}} {b}−\mathrm{2}{Ab}^{\mathrm{3}} \right){x}−\left(\mathrm{8}{Aa}^{\mathrm{2}} {b}^{\mathrm{3}} −{a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right){y}−\mathrm{2}{ab}\left({a}^{\mathrm{3}} +{ab}^{\mathrm{2}} −\mathrm{2}{Aab}^{\mathrm{3}} \right)=\mathrm{0} \\ $$
Commented by mrW2 last updated on 15/Jan/18
Commented by ajfour last updated on 15/Jan/18
$${Great}\:{Sir}!{this}\:{is}\:{truly}\:{amazing}. \\ $$$${i}\:{shall}\:{try}\:{to}\:{follow},\:{might}\:{take}\:{a}\:{while}. \\ $$
Commented by mrW2 last updated on 15/Jan/18
Commented by mrW2 last updated on 15/Jan/18
$${This}\:{is}\:{also}\:{a}\:{good}\:{way}\:{in}\:{current}\:{case}, \\ $$$${because}\:{the}\:{curve}\:{to}\:{reflect}\:{is}\:{a}\: \\ $$$${symmetric}\:{figure}.\:{If}\:{the}\:{figure}\:{is} \\ $$$${not}\:{symmetric},\:{e}.{g}.\:{if}\:{y}={e}^{{x}} ,\:{you}\:{can} \\ $$$${not}\:{obtain}\:{the}\:{mirrored}\:{function} \\ $$$${through}\:{such}\:{a}\:{rotation}. \\ $$
Commented by mrW2 last updated on 16/Jan/18