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Question-204168

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Question Number 204168 by mr W last updated on 07/Feb/24
Commented by mr W last updated on 07/Feb/24
find the locus of point P.
$${find}\:{the}\:{locus}\:{of}\:{point}\:{P}. \\ $$
Answered by mr W last updated on 08/Feb/24
say P(u,v) line through point P: y=m(x−u)+v=mx+v−mu due to tangency: m^2 a^2 +b^2 =(v−mu)^2 (a^2 −u^2 )m^2 +2uvm+b^2 −v^2 =0 m_1 +m_2 =−((2uv)/(a^2 −u^2 )) m_1 m_2 =((b^2 −v^2 )/(a^2 −u^2 )) m_2 −m_1 =±(√((m_2 +m_1 )^2 −4m_2 m_1 )) =±((2(√(a^2 v^2 +b^2 u^2 −a^2 b^2 )))/(a^2 −u^2 )) m_1 =tan θ_1 m_2 =tan θ_2 θ_2 −θ_1 =θ=(π/3) ((tan θ_2 −tan θ_1 )/(1+tan θ_2 tan θ_1 ))=tan θ=k, say ((m_2 −m_1 )/(1+m_2 m_1 ))=k ((2(√(a^2 v^2 +b^2 u^2 −a^2 b^2 )))/((a^2 −u^2 )(1+((b^2 −v^2 )/(a^2 −u^2 )))))=±k ⇒((2(√(a^2 v^2 +b^2 u^2 −a^2 b^2 )))/(a^2 +b^2 −u^2 −v^2 ))=±k locus of point P is ((2(√(a^2 y^2 +b^2 x^2 −a^2 b^2 )))/(a^2 +b^2 −x^2 −y^2 ))=±tan θ if θ=90°, a^2 +b^2 −x^2 −y^2 =0 ⇒circle with r=(√(a^2 +b^2 ))
$${say}\:{P}\left({u},{v}\right) \\ $$$${line}\:{through}\:{point}\:{P}: \\ $$$${y}={m}\left({x}−{u}\right)+{v}={mx}+{v}−{mu} \\ $$$${due}\:{to}\:{tangency}: \\ $$$${m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({v}−{mu}\right)^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{u}^{\mathrm{2}} \right){m}^{\mathrm{2}} +\mathrm{2}{uvm}+{b}^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}_{\mathrm{1}} +{m}_{\mathrm{2}} =−\frac{\mathrm{2}{uv}}{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −{v}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${m}_{\mathrm{2}} −{m}_{\mathrm{1}} =\pm\sqrt{\left({m}_{\mathrm{2}} +{m}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{4}{m}_{\mathrm{2}} {m}_{\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:=\pm\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${m}_{\mathrm{1}} =\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${m}_{\mathrm{2}} =\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$$\theta_{\mathrm{2}} −\theta_{\mathrm{1}} =\theta=\frac{\pi}{\mathrm{3}} \\ $$$$\frac{\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} }{\mathrm{1}+\mathrm{tan}\:\theta_{\mathrm{2}} \mathrm{tan}\:\theta_{\mathrm{1}} }=\mathrm{tan}\:\theta={k},\:{say} \\ $$$$\frac{{m}_{\mathrm{2}} −{m}_{\mathrm{1}} }{\mathrm{1}+{m}_{\mathrm{2}} {m}_{\mathrm{1}} }={k} \\ $$$$\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}{\left({a}^{\mathrm{2}} −{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} −{v}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} }\right)}=\pm{k} \\ $$$$\Rightarrow\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{u}^{\mathrm{2}} −{v}^{\mathrm{2}} }=\pm{k} \\ $$$${locus}\:{of}\:{point}\:{P}\:{is} \\ $$$$\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {y}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }=\pm\mathrm{tan}\:\theta \\ $$$${if}\:\theta=\mathrm{90}°, \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{circle}\:{with}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$

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