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Question Number 77127 by peter frank last updated on 03/Jan/20

Prove that line lx+my+n=0 is tangent to the ellipse (x^2 /a^2 )+(y^2 /b^(2 ) )=1 if a^2 l^2 +b^2 m^2 =n^2

$${Prove}\:{that}\:{line}\:{lx}+{my}+{n}=\mathrm{0} \\ $$$${is}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}\:} }=\mathrm{1}\:{if}\:{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$

Answered by jagoll last updated on 03/Jan/20

suppose a > b ⇔ tangent line ellips (x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ y = mx ± (√(a^2 m^2 +b^2 )) equal to lx +my +n = 0⇒ y = −(l/m)x−(n/m) m = −(l/m) ⇒ m^2 =−l −(n/m)=−(√(a^2 m^2 +b^2 )) ⇒ (n^2 /m^2 )=a^2 m^2 +b^2 n^2 =(−l)(−la^2 +b^2 ) n^2 = l^2 a^2 −lb^2 ⇒n^2 =l^2 a^2 +m^2 b^2

$${suppose}\:{a}\:>\:{b}\:\Leftrightarrow\:{tangent}\:{line}\: \\ $$$${ellips}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\:{y}\:=\:{mx}\:\pm\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${equal}\:{to}\:{lx}\:+{my}\:+{n}\:=\:\mathrm{0}\Rightarrow\: \\ $$$${y}\:=\:−\frac{{l}}{{m}}{x}−\frac{{n}}{{m}}\: \\ $$$${m}\:=\:−\frac{{l}}{{m}}\:\Rightarrow\:{m}^{\mathrm{2}} =−{l}\: \\ $$$$−\frac{{n}}{{m}}=−\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{{n}^{\mathrm{2}} }{{m}^{\mathrm{2}} }={a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \: \\ $$$${n}^{\mathrm{2}} =\left(−{l}\right)\left(−{la}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\: \\ $$$${n}^{\mathrm{2}} =\:{l}^{\mathrm{2}} {a}^{\mathrm{2}} −{lb}^{\mathrm{2}} \:\Rightarrow{n}^{\mathrm{2}} ={l}^{\mathrm{2}} {a}^{\mathrm{2}} +{m}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$

Commented by peter frank last updated on 03/Jan/20

thank you

$${thank}\:{you} \\ $$

Answered by mr W last updated on 03/Jan/20

b^2 x^2 +a^2 y^2 =a^2 b^2 m^2 b^2 x^2 +a^2 m^2 y^2 =m^2 a^2 b^2 m^2 b^2 x^2 +a^2 (lx+n)^2 =m^2 a^2 b^2 m^2 b^2 x^2 +a^2 (l^2 x^2 +n^2 +2lnx)=m^2 a^2 b^2 (a^2 l^2 +b^2 m^2 )x^2 +2a^2 lnx+a^2 (n^2 −b^2 m^2 )=0 due to tangency: Δ=(2a^2 ln)^2 −4(a^2 l^2 +b^2 m^2 )a^2 (n^2 −b^2 m^2 )=0 a^2 l^2 n^2 −(a^2 l^2 +b^2 m^2 )(n^2 −b^2 m^2 )=0 (−n^2 +a^2 l^2 +b^2 m^2 )b^2 m^2 =0 ⇒a^2 l^2 +b^2 m^2 =n^2

$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {m}^{\mathrm{2}} {y}^{\mathrm{2}} ={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({lx}+{n}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({l}^{\mathrm{2}} {x}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{lnx}\right)={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {lnx}+{a}^{\mathrm{2}} \left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${due}\:{to}\:{tangency}: \\ $$$$\Delta=\left(\mathrm{2}{a}^{\mathrm{2}} {ln}\right)^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){a}^{\mathrm{2}} \left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {l}^{\mathrm{2}} {n}^{\mathrm{2}} −\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)\left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left(−{n}^{\mathrm{2}} +{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){b}^{\mathrm{2}} {m}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$

Commented by peter frank last updated on 03/Jan/20

thank you

$${thank}\:{you} \\ $$

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