# 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-

Question Number 77127 by peter frank last updated on 03/Jan/20
$${Prove}\:{that}\:{line}\:{lx}+{my}+{n}=\mathrm{0} \\$$$${is}\:{tangent}\:{to}\:{the}\:{ellipse} \\$$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}\:} }=\mathrm{1}\:{if}\:{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\$$
Answered by jagoll last updated on 03/Jan/20
$${suppose}\:{a}\:>\:{b}\:\Leftrightarrow\:{tangent}\:{line}\: \\$$$${ellips}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\:{y}\:=\:{mx}\:\pm\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\$$$${equal}\:{to}\:{lx}\:+{my}\:+{n}\:=\:\mathrm{0}\Rightarrow\: \\$$$${y}\:=\:−\frac{{l}}{{m}}{x}−\frac{{n}}{{m}}\: \\$$$${m}\:=\:−\frac{{l}}{{m}}\:\Rightarrow\:{m}^{\mathrm{2}} =−{l}\: \\$$$$−\frac{{n}}{{m}}=−\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\$$$$\Rightarrow\:\frac{{n}^{\mathrm{2}} }{{m}^{\mathrm{2}} }={a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \: \\$$$${n}^{\mathrm{2}} =\left(−{l}\right)\left(−{la}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\: \\$$$${n}^{\mathrm{2}} =\:{l}^{\mathrm{2}} {a}^{\mathrm{2}} −{lb}^{\mathrm{2}} \:\Rightarrow{n}^{\mathrm{2}} ={l}^{\mathrm{2}} {a}^{\mathrm{2}} +{m}^{\mathrm{2}} {b}^{\mathrm{2}} \\$$
Commented by peter frank last updated on 03/Jan/20
$${thank}\:{you} \\$$
Answered by mr W last updated on 03/Jan/20
$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\$$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {m}^{\mathrm{2}} {y}^{\mathrm{2}} ={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\$$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({lx}+{n}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\$$$${m}^{\mathrm{2}} {b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({l}^{\mathrm{2}} {x}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{lnx}\right)={m}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\$$$$\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {lnx}+{a}^{\mathrm{2}} \left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\$$$${due}\:{to}\:{tangency}: \\$$$$\Delta=\left(\mathrm{2}{a}^{\mathrm{2}} {ln}\right)^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){a}^{\mathrm{2}} \left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\$$$${a}^{\mathrm{2}} {l}^{\mathrm{2}} {n}^{\mathrm{2}} −\left({a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)\left({n}^{\mathrm{2}} −{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0} \\$$$$\left(−{n}^{\mathrm{2}} +{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right){b}^{\mathrm{2}} {m}^{\mathrm{2}} =\mathrm{0} \\$$$$\Rightarrow{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\$$
Commented by peter frank last updated on 03/Jan/20
$${thank}\:{you} \\$$