Coordinate Geometry Questions

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

Commented by ajfour last updated on 27/Jul/20

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

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

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

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