Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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.

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

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!

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

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

i made a mistake. now it should be correct.

$${i}\:{made}\:{a}\:{mistake}.\:{now}\:{it}\:{should}\:{be} \\ $$$${correct}. \\ $$

Commented by ajfour last updated on 13/Feb/20

Beautiful ! how come the blue curves dont meet the x-axis, Sir?

$${Beautiful}\:! \\ $$$${how}\:{come}\:{the}\:{blue}\:{curves}\:{dont} \\ $$$${meet}\:{the}\:{x}-{axis},\:{Sir}? \\ $$$$ \\ $$

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

Commented by ajfour last updated on 14/Feb/20

thanks sir, for taking my remark seriously, now it seems more natural type..

$${thanks}\:{sir},\:{for}\:{taking}\:{my} \\ $$$${remark}\:{seriously},\:{now}\:{it} \\ $$$${seems}\:{more}\:{natural}\:{type}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com