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

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