Question Number 43331 by ajfour last updated on 09/Sep/18
Commented by ajfour last updated on 09/Sep/18
$${Find}\:{equation}\:{of}\:{red}\:{parabola}, \\ $$$${if}\:{it}\:{be}\:{reflection}\:{of}\:{y}={x}^{\mathrm{2}} \:\:{in} \\ $$$${the}\:{line}\:{y}={mx}+{c}\:. \\ $$
Commented by ajfour last updated on 09/Sep/18
Commented by ajfour last updated on 09/Sep/18
![p=((c+mh−k)/(√(1+m^2 ))) y−k = 2pcos θ h−x = 2psin θ tan 𝛉 = m ⇒ x=h−((2m(c+mh−k))/(1+m^2 )) and y= k+((2(c+mh−k))/(1+m^2 )) Lets consider a point P(h,k) of red parabola. ⇒ f(h,k)=0 let image of P (h,k) be I(x,y) But y=x^2 , ⇒ k+((2(c+mh−k))/(1+m^2 )) =[ h−((2m(c+mh−k))/(1+m^2 ))]^2 hence f(x,y) is y+((2(c+mx−y))/(1+m^2 ))=[x−((2m(c+mx−y))/(1+m^2 ))]^2 . Just to check let m=1 , c=0 ⇒ x=y^2 (reflection of y=x^2 ) .](Q43336.png)
$${p}=\frac{{c}+{mh}−{k}}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$${y}−{k}\:=\:\mathrm{2}{p}\mathrm{cos}\:\theta \\ $$$${h}−{x}\:=\:\mathrm{2}{p}\mathrm{sin}\:\theta \\ $$$$\mathrm{tan}\:\boldsymbol{\theta}\:=\:\boldsymbol{{m}} \\ $$$$\Rightarrow\:\boldsymbol{{x}}=\boldsymbol{{h}}−\frac{\mathrm{2}\boldsymbol{{m}}\left(\boldsymbol{{c}}+\boldsymbol{{mh}}−\boldsymbol{{k}}\right)}{\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} } \\ $$$${and}\:\:\:\boldsymbol{{y}}=\:\boldsymbol{{k}}+\frac{\mathrm{2}\left(\boldsymbol{{c}}+\boldsymbol{{mh}}−\boldsymbol{{k}}\right)}{\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} } \\ $$$${Lets}\:{consider}\:{a}\:{point}\:{P}\left({h},{k}\right)\:{of} \\ $$$${red}\:{parabola}. \\ $$$$\Rightarrow\:\:{f}\left({h},{k}\right)=\mathrm{0} \\ $$$${let}\:{image}\:{of}\:{P}\:\left({h},{k}\right)\:{be}\:{I}\left({x},{y}\right) \\ $$$${But}\:\:{y}={x}^{\mathrm{2}} \:\:\:\:,\:\Rightarrow \\ $$$$\:\boldsymbol{{k}}+\frac{\mathrm{2}\left(\boldsymbol{{c}}+\boldsymbol{{mh}}−\boldsymbol{{k}}\right)}{\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} }\:=\left[\:\boldsymbol{{h}}−\frac{\mathrm{2}\boldsymbol{{m}}\left(\boldsymbol{{c}}+\boldsymbol{{mh}}−\boldsymbol{{k}}\right)}{\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} }\right]^{\mathrm{2}} \\ $$$${hence}\:{f}\left({x},{y}\right)\:{is} \\ $$$${y}+\frac{\mathrm{2}\left({c}+{mx}−{y}\right)}{\mathrm{1}+{m}^{\mathrm{2}} }=\left[{x}−\frac{\mathrm{2}{m}\left({c}+{mx}−{y}\right)}{\mathrm{1}+{m}^{\mathrm{2}} }\right]^{\mathrm{2}} . \\ $$$${Just}\:{to}\:{check} \\ $$$${let}\:{m}=\mathrm{1}\:,\:{c}=\mathrm{0} \\ $$$$\Rightarrow\:{x}={y}^{\mathrm{2}} \:\:\:\:\left({reflection}\:{of}\:{y}={x}^{\mathrm{2}} \right)\:. \\ $$
Commented by ajfour last updated on 10/Sep/18
$${Thank}\:{you}\:{Sir}. \\ $$
Commented by MrW3 last updated on 09/Sep/18
$${correct}\:{sir}. \\ $$$${Generally}\:{mirror}\:{image}\:{of}\:{point}\:{P}\left({x},{y}\right)\:{in} \\ $$$${line}\:\boldsymbol{{ax}}+\boldsymbol{{by}}+\boldsymbol{{c}}=\mathrm{0}\:{is}\:{point}\:{Q}\left({u},{v}\right){with} \\ $$$${u}={x}−\frac{\mathrm{2}{a}\left({ax}+{by}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${v}={y}−\frac{\mathrm{2}{b}\left({ax}+{by}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$