Prove-that-the-locus-of-the-point-of-intersection-of-perpendicular-tangents-to-an-ellipse-is-another-ellipse-

Question Number 81470 by ajfour last updated on 13/Feb/20

$${Prove}\:{that}\:{the}\:{locus}\:{of}\:{the}\:{point} \\ $$$${of}\:{intersection}\:{of}\:{perpendicular} \\ $$$${tangents}\:{to}\:{an}\:{ellipse}\:{is}\:{another} \\ $$$${ellipse}. \\ $$

Answered by mr W last updated on 13/Feb/20

say the ellipse is (x^2 /a^2 )+(y^2 /b^2 )=1 say the intersection of perpendicular tangents is P(u,v). that means from point P(u,v) the two tangents are perpendicular to each other. say the tangent from P(u,v) is y=v+m(x−u) or mx−y=mu−v we have a^2 m^2 +b^2 =(mu−v)^2 (a^2 −u^2 )m^2 +2uvm+b^2 −v^2 =0 ⇒m_1 m_2 =((b^2 −v^2 )/(a^2 −u^2 )) since both tangents are ⊥ to each other, m_1 m_2 =−1 ⇒=((b^2 −v^2 )/(a^2 −u^2 ))=−1 ⇒u^2 +v^2 =a^2 +b^2 =((√(a^2 +b^2 )))^2 i.e. the locus of point P is: x^2 +y^2 =((√(a^2 +b^2 )))^2 it is a circle (special ellipse) with the same center as the ellipse and radius of (√(a^2 +b^2 )).

$${say}\:{the}\:{ellipse}\:{is} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${say}\:{the}\:{intersection}\:{of}\:{perpendicular} \\ $$$${tangents}\:{is}\:{P}\left({u},{v}\right).\:{that}\:{means}\:{from} \\ $$$${point}\:{P}\left({u},{v}\right)\:{the}\:{two}\:{tangents}\:{are} \\ $$$${perpendicular}\:{to}\:{each}\:{other}. \\ $$$${say}\:{the}\:{tangent}\:{from}\:{P}\left({u},{v}\right)\:{is} \\ $$$${y}={v}+{m}\left({x}−{u}\right)\:{or} \\ $$$${mx}−{y}={mu}−{v} \\ $$$${we}\:{have} \\ $$$${a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({mu}−{v}\right)^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{u}^{\mathrm{2}} \right){m}^{\mathrm{2}} +\mathrm{2}{uvm}+{b}^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −{v}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${since}\:{both}\:{tangents}\:{are}\:\bot\:{to}\:{each}\:{other}, \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\Rightarrow=\frac{{b}^{\mathrm{2}} −{v}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} }=−\mathrm{1} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +{v}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${i}.{e}.\:{the}\:{locus}\:{of}\:{point}\:{P}\:{is}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${it}\:{is}\:{a}\:{circle}\:\left({special}\:{ellipse}\right)\:{with}\:{the} \\ $$$${same}\:{center}\:{as}\:{the}\:{ellipse}\:{and}\:{radius} \\ $$$${of}\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }. \\ $$

Commented by mr W last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

$${Thanks}\:{Sir},\:{but}\:{thats}\:{how}\:{the} \\ $$$${question}\:{was}\:{written}\:{in}\:{a}\:{book}! \\ $$

Commented by mr W last updated on 13/Feb/20

if the tangents are not ⊥ to each other, but build an angle θ, then we have tan θ=tan (θ_1 −θ_2 )=((m_1 −m_2 )/(1+m_1 m_2 )) m_1 +m_2 =−((2uv)/(a^2 −u^2 )) m_1 m_2 =((b^2 −v^2 )/(a^2 −u^2 )) (m_1 −m_2 )^2 =(m_1 +m_2 )^2 −4m_1 m_2 =((4(a^2 v^2 +b^2 u^2 −a^2 b^2 ))/((a^2 −u^2 )^2 )) tan^2 θ=(((m_1 −m_2 )^2 )/((1+m_1 m_2 )^2 ))=((4(a^2 v^2 +b^2 u^2 −a^2 b^2 ))/((a^2 −u^2 +b^2 −v^2 )^2 )) tan^2 θ (u^2 +v^2 −a^2 −b^2 )^2 =4(a^2 v^2 +b^2 u^2 −a^2 b^2 ) or tan^2 θ (x^2 +y^2 −a^2 −b^2 )^2 =4(b^2 x^2 +a^2 y^2 −a^2 b^2 ) example: θ=60°

$${if}\:{the}\:{tangents}\:{are}\:{not}\:\bot\:{to}\:{each}\:{other}, \\ $$$${but}\:{build}\:{an}\:{angle}\:\theta,\:{then}\:{we}\:{have} \\ $$$$\mathrm{tan}\:\theta=\mathrm{tan}\:\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)=\frac{{m}_{\mathrm{1}} −{m}_{\mathrm{2}} }{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} } \\ $$$${m}_{\mathrm{1}} +{m}_{\mathrm{2}} =−\frac{\mathrm{2}{uv}}{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −{v}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$$\left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{m}_{\mathrm{1}} {m}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}\left({a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} −{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta=\frac{\left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{4}\left({a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} −{u}^{\mathrm{2}} +{b}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta\:\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}\left({a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right) \\ $$$${or} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}\left({b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right) \\ $$$$ \\ $$$${example}:\:\theta=\mathrm{60}° \\ $$

Commented by mr W last updated on 13/Feb/20