Question-105273

Question Number 105273 by ajfour last updated on 27/Jul/20

Commented by ajfour last updated on 27/Jul/20

Given P(h,k) and B(p,q) To find image I(r,t) of B(p,q) in the mirror line AT.

$${Given}\:{P}\left({h},{k}\right)\:\:\:{and}\:\:{B}\left({p},{q}\right) \\ $$$${To}\:{find}\:{image}\:{I}\left({r},{t}\right)\:{of}\:{B}\left({p},{q}\right) \\ $$$${in}\:{the}\:{mirror}\:{line}\:{AT}. \\ $$

Answered by mr W last updated on 28/Jul/20

T(m,m^2 ) tan θ=((b−m^2 )/(a−m))=2m m^2 −2am+b=0 ⇒m=a+(√(a^2 −b)) eqn. of AP: y=b+2m(x−a) ⇒2mx−y+b−2ma image of B(p,q) is (r,t): r=p−((4m(2mp−q+b−2ma))/(4m^2 +1)) t=q+((2(2mp−q+b−2ma))/(4m^2 +1)) image of P(h,k) is (f,g): f=h−((4m(2mh−k+b−2ma))/(4m^2 +1)) g=k+((2(2mh−k+b−2ma))/(4m^2 +1))

$${T}\left({m},{m}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\frac{{b}−{m}^{\mathrm{2}} }{{a}−{m}}=\mathrm{2}{m} \\ $$$${m}^{\mathrm{2}} −\mathrm{2}{am}+{b}=\mathrm{0} \\ $$$$\Rightarrow{m}={a}+\sqrt{{a}^{\mathrm{2}} −{b}} \\ $$$${eqn}.\:{of}\:{AP}: \\ $$$${y}={b}+\mathrm{2}{m}\left({x}−{a}\right) \\ $$$$\Rightarrow\mathrm{2}{mx}−{y}+{b}−\mathrm{2}{ma} \\ $$$$ \\ $$$${image}\:{of}\:{B}\left({p},{q}\right)\:{is}\:\left({r},{t}\right): \\ $$$${r}={p}−\frac{\mathrm{4}{m}\left(\mathrm{2}{mp}−{q}+{b}−\mathrm{2}{ma}\right)}{\mathrm{4}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$${t}={q}+\frac{\mathrm{2}\left(\mathrm{2}{mp}−{q}+{b}−\mathrm{2}{ma}\right)}{\mathrm{4}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$${image}\:{of}\:{P}\left({h},{k}\right)\:{is}\:\left({f},{g}\right): \\ $$$${f}={h}−\frac{\mathrm{4}{m}\left(\mathrm{2}{mh}−{k}+{b}−\mathrm{2}{ma}\right)}{\mathrm{4}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$${g}={k}+\frac{\mathrm{2}\left(\mathrm{2}{mh}−{k}+{b}−\mathrm{2}{ma}\right)}{\mathrm{4}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by mr W last updated on 28/Jul/20

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