Menu Close

Question-66161

Tinku Tara 0 Comments
Pin




Question Number 66161 by peter frank last updated on 09/Aug/19
Answered by mr W last updated on 10/Aug/19
if a line y=mx+c tangents the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 we have (x^2 /a^2 )+(((mx+c)^2 )/b^2 )=1 b^2 x^2 +a^2 (m^2 x^2 +2mcx+c^2 )=a^2 b^2 (m^2 a^2 +b^2 )x^2 +2mca^2 x+a^2 (c^2 −b^2 )=0 Δ=4m^2 c^2 a^4 −4(m^2 a^2 +b^2 )a^2 (c^2 −b^2 )=0 m^2 c^2 a^2 −(m^2 a^2 +b^2 )(c^2 −b^2 )=0 ⇒m^2 a^2 +b^2 =c^2 let′s say all the three ellipses have a common tangent line y=mx+c, then m^2 a_1 ^2 +b_1 ^2 =c^2 ...(i) m^2 a_2 ^2 +b_2 ^2 =c^2 ...(ii) m^2 a_3 ^2 +b_3 ^2 =c^2 ...(iii) (i)−(ii): m^2 (a_1 ^2 −a_2 ^2 )+(b_1 ^2 −b_2 ^2 )=0 ...(iv) (i)−(iii): m^2 (a_1 ^2 −a_3 ^2 )+(b_1 ^2 −b_3 ^2 )=0 ...(v) from (iv) and (v): (a_1 ^2 −a_3 ^2 )(b_1 ^2 −b_2 ^2 )−(a_1 ^2 −a_2 ^2 )(b_1 ^2 −b_3 ^2 )=0 a_1 ^2 b_1 ^2 −a_3 ^2 b_1 ^2 −a_1 ^2 b_2 ^2 +a_3 ^2 b_2 ^2 −a_1 ^2 b_1 ^2 +a_2 ^2 b_1 ^2 +a_1 ^2 b_3 ^2 −a_2 ^2 b_3 ^2 =0 ⇒(a_2 ^2 b_3 ^2 −a_3 ^2 b_2 ^2 )−(a_1 ^2 b_3 ^2 −a_3 ^2 b_1 ^2 )+(a_1 ^2 b_2 ^2 −a_2 ^2 b_1 ^2 )=0 ⇒ determinant ((a_1 ^2 ,b_1 ^2 ,1),(a_2 ^2 ,b_2 ^2 ,1),(a_3 ^2 ,b_3 ^2 ,1))=0
$${if}\:{a}\:{line}\:{y}={mx}+{c}\:{tangents}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${we}\:{have} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({mx}+{c}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({m}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{mcx}+{c}^{\mathrm{2}} \right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{mca}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Delta=\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} {a}^{\mathrm{4}} −\mathrm{4}\left({m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${m}^{\mathrm{2}} {c}^{\mathrm{2}} {a}^{\mathrm{2}} −\left({m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$ \\ $$$${let}'{s}\:{say}\:{all}\:{the}\:{three}\:{ellipses}\:{have}\:{a} \\ $$$${common}\:{tangent}\:{line}\:{y}={mx}+{c},\:{then} \\ $$$${m}^{\mathrm{2}} {a}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$${m}^{\mathrm{2}} {a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:\:…\left({ii}\right) \\ $$$${m}^{\mathrm{2}} {a}_{\mathrm{3}} ^{\mathrm{2}} +{b}_{\mathrm{3}} ^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${m}^{\mathrm{2}} \left({a}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{2}} ^{\mathrm{2}} \right)+\left({b}_{\mathrm{1}} ^{\mathrm{2}} −{b}_{\mathrm{2}} ^{\mathrm{2}} \right)=\mathrm{0}\:\:\:…\left({iv}\right) \\ $$$$\left({i}\right)−\left({iii}\right): \\ $$$${m}^{\mathrm{2}} \left({a}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{3}} ^{\mathrm{2}} \right)+\left({b}_{\mathrm{1}} ^{\mathrm{2}} −{b}_{\mathrm{3}} ^{\mathrm{2}} \right)=\mathrm{0}\:\:\:…\left({v}\right) \\ $$$${from}\:\left({iv}\right)\:{and}\:\left({v}\right): \\ $$$$\left({a}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{3}} ^{\mathrm{2}} \right)\left({b}_{\mathrm{1}} ^{\mathrm{2}} −{b}_{\mathrm{2}} ^{\mathrm{2}} \right)−\left({a}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{2}} ^{\mathrm{2}} \right)\left({b}_{\mathrm{1}} ^{\mathrm{2}} −{b}_{\mathrm{3}} ^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} −{a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{2}} −{a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} −{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} −{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{2}} \right)−\left({a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} −{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} \right)+\left({a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{2}} −{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\begin{vmatrix}{{a}_{\mathrm{1}} ^{\mathrm{2}} }&{{b}_{\mathrm{1}} ^{\mathrm{2}} }&{\mathrm{1}}\\{{a}_{\mathrm{2}} ^{\mathrm{2}} }&{{b}_{\mathrm{2}} ^{\mathrm{2}} }&{\mathrm{1}}\\{{a}_{\mathrm{3}} ^{\mathrm{2}} }&{{b}_{\mathrm{3}} ^{\mathrm{2}} }&{\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$
Commented by peter frank last updated on 10/Aug/19
thank you very much
$${thank}\:{you}\:{very}\:{much}\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *