Question Number 59870 by ajfour last updated on 15/May/19
Commented by ajfour last updated on 15/May/19
$$\mathrm{Find}\:\mathrm{minimum}\:\mathrm{length}\:\mathrm{of}\:\mathrm{AB}.\:\mathrm{A}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{an}\:\mathrm{ellipse}\:\mathrm{while}\:\mathrm{B}\:\mathrm{on}\:\mathrm{shown}\:\mathrm{parabola}. \\ $$
Commented by MJS last updated on 15/May/19
$$\mathrm{am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{or}\:\mathrm{wrong}? \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{parallel}\:\mathrm{tangents}\:\mathrm{with}\:\mathrm{0}<{p},\:{q}<\mathrm{2}{a} \\ $$$$\mathrm{with}\:\mathrm{maximum}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{those} \\ $$$$\mathrm{tangents} \\ $$
Commented by mr W last updated on 15/May/19
$${you}\:{are}\:{right}\:{sir}. \\ $$$${but}:\:{with}\:{minimum}\:{distance}\:{between}… \\ $$
Commented by MJS last updated on 16/May/19
![I tried this: let a=1 [it′s just a scaling factor] P∈par= ((p),((f_1 (p))) ) Q∈ell= ((q),((f_2 (q))) ) min. distance=min(∣PQ∣)=min(∣PQ∣^2 ) D=∣PQ∣^2 =(p^4 /(16))−(p^3 /2)+3p^2 −4p−2pq+((3q^2 )/4)+5−((p^2 /4)−p+2)(√(4−q^2 )) (dD/dp)=(p^3 /4)−((3p^2 )/2)+6p−2q−4−((p/2)−1)(√(4−q^2 )) ⇒ p^3 −6p^2 +2p(12−(√(4−q^2 )))−4(2(q+2)−(√(4−q^2 )))=0 p=t+2 t^3 +2(6−(√(4−q^2 )))t+8(2−q)=0 this has got 1 real and 2 complex solutions ⇒ t_1 =u^(1/3) −v^(1/3) ⇒ p_1 =2+u^(1/3) −v^(1/3) u^(1/3) =((4q−8+(2/9)(√(6(36(q^2 −6q+14)+(q^2 −112(√(4−q^2 )))))))^(1/3) v^(1/3) =((8−4q+(2/9)(√(6(36(q^2 −6q+14)+(q^2 −112(√(4−q^2 )))))))^(1/3) now we′d have to insert p_1 in D and find the minimum of (dD/dq) I approximated with my TI−89 graphic calculator and got this: q=1.167275 p=1.281246 P= (((1.281246)),((1.129152)) ) Q= (((1.167275)),((.8120145)) ) D=.3369949](https://www.tinkutara.com/question/Q59964.png)
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}: \\ $$$$\mathrm{let}\:{a}=\mathrm{1}\:\left[\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{scaling}\:\mathrm{factor}\right] \\ $$$${P}\in\mathrm{par}=\begin{pmatrix}{{p}}\\{{f}_{\mathrm{1}} \left({p}\right)}\end{pmatrix}\:\:{Q}\in\mathrm{ell}=\begin{pmatrix}{{q}}\\{{f}_{\mathrm{2}} \left({q}\right)}\end{pmatrix} \\ $$$$\mathrm{min}.\:\mathrm{distance}=\mathrm{min}\left(\mid{PQ}\mid\right)=\mathrm{min}\left(\mid{PQ}\mid^{\mathrm{2}} \right) \\ $$$${D}=\mid{PQ}\mid^{\mathrm{2}} =\frac{{p}^{\mathrm{4}} }{\mathrm{16}}−\frac{{p}^{\mathrm{3}} }{\mathrm{2}}+\mathrm{3}{p}^{\mathrm{2}} −\mathrm{4}{p}−\mathrm{2}{pq}+\frac{\mathrm{3}{q}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{5}−\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{p}+\mathrm{2}\right)\sqrt{\mathrm{4}−{q}^{\mathrm{2}} } \\ $$$$\frac{{dD}}{{dp}}=\frac{{p}^{\mathrm{3}} }{\mathrm{4}}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6}{p}−\mathrm{2}{q}−\mathrm{4}−\left(\frac{{p}}{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{4}−{q}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{3}} −\mathrm{6}{p}^{\mathrm{2}} +\mathrm{2}{p}\left(\mathrm{12}−\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\right)−\mathrm{4}\left(\mathrm{2}\left({q}+\mathrm{2}\right)−\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$${p}={t}+\mathrm{2} \\ $$$${t}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{6}−\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\right){t}+\mathrm{8}\left(\mathrm{2}−{q}\right)=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\Rightarrow \\ $$$${t}_{\mathrm{1}} ={u}^{\mathrm{1}/\mathrm{3}} −{v}^{\mathrm{1}/\mathrm{3}} \:\Rightarrow\:{p}_{\mathrm{1}} =\mathrm{2}+{u}^{\mathrm{1}/\mathrm{3}} −{v}^{\mathrm{1}/\mathrm{3}} \\ $$$${u}^{\mathrm{1}/\mathrm{3}} =\sqrt[{\mathrm{3}}]{\mathrm{4}{q}−\mathrm{8}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{6}\left(\mathrm{36}\left({q}^{\mathrm{2}} −\mathrm{6}{q}+\mathrm{14}\right)+\left({q}^{\mathrm{2}} −\mathrm{112}\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\right)\right.}} \\ $$$${v}^{\mathrm{1}/\mathrm{3}} =\sqrt[{\mathrm{3}}]{\mathrm{8}−\mathrm{4}{q}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{6}\left(\mathrm{36}\left({q}^{\mathrm{2}} −\mathrm{6}{q}+\mathrm{14}\right)+\left({q}^{\mathrm{2}} −\mathrm{112}\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\right)\right.}} \\ $$$$\mathrm{now}\:\mathrm{we}'\mathrm{d}\:\mathrm{have}\:\mathrm{to}\:\mathrm{insert}\:{p}_{\mathrm{1}} \:\mathrm{in}\:{D}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{minimum}\:\mathrm{of}\:\frac{{dD}}{{dq}} \\ $$$$\mathrm{I}\:\mathrm{approximated}\:\mathrm{with}\:\mathrm{my}\:\mathrm{TI}−\mathrm{89}\:\mathrm{graphic} \\ $$$$\mathrm{calculator}\:\mathrm{and}\:\mathrm{got}\:\mathrm{this}: \\ $$$${q}=\mathrm{1}.\mathrm{167275} \\ $$$${p}=\mathrm{1}.\mathrm{281246} \\ $$$${P}=\begin{pmatrix}{\mathrm{1}.\mathrm{281246}}\\{\mathrm{1}.\mathrm{129152}}\end{pmatrix}\:\:{Q}=\begin{pmatrix}{\mathrm{1}.\mathrm{167275}}\\{.\mathrm{8120145}}\end{pmatrix} \\ $$$${D}=.\mathrm{3369949} \\ $$
Commented by mr W last updated on 16/May/19
$${you}\:{got}\:{correct}\:{result}\:{sir}! \\ $$
Commented by ajfour last updated on 16/May/19
$$\mathrm{awesome}\:\mathrm{effort},\:\mathrm{thanks}\:\mathrm{sir}! \\ $$
Commented by ajfour last updated on 16/May/19
$$\mathrm{yes}\://\mathrm{tangents}\:\mathrm{with}\:\mathrm{a}\:\left(\mathrm{local}\right)\:\mathrm{maximum} \\ $$$$\mathrm{distance}. \\ $$
Answered by mr W last updated on 16/May/19
![method 1: eqn. of parabola: y=(((x−2a)^2 )/(4a))+a eqn. of ellipse: (x^2 /(4a^2 ))+(y^2 /a^2 )=1 A(p,c) B(q,d) (p^2 /(4a^2 ))+(c^2 /a^2 )=1 ⇒c=a(√(1−(1/4)((p/a))^2 )) ⇒c=(a/2)(√(4−λ^2 )) d=(1/(4a))(q−2a)^2 +a=a[1+(1/4)((q/a)−2)^2 ] ⇒d=(a/4)[4+(μ−2)^2 ] D=(AB)^2 =(p−q)^2 +(c−d)^2 =a^2 [(λ−μ)^2 +{(1/2)(√(4−λ^2 ))−1−(((μ−2)^2 )/4)}^2 ]=a^2 F(λ,μ) F(λ,μ)=(λ−μ)^2 +{(1/2)(√(4−λ^2 ))−1−(((μ−2)^2 )/4)}^2 (∂F/∂λ)=2(λ−μ)+2{(1/2)(√(4−λ^2 ))−1−(((μ−2)^2 )/4)}(1/(4(√(4−λ^2 ))))×(−2λ)=0 ⇒4(λ−μ)(√(4−λ^2 ))=λ{(√(4−λ^2 ))−2−(((μ−2)^2 )/2)} ...(i) (∂F/∂μ)=−2(λ−μ)+2{(1/2)(√(4−λ^2 ))−1−(((μ−2)^2 )/4)}{−((2(μ−2))/4)}=0 ⇒4(λ−μ)=(2−μ){(√(4−λ^2 ))−2−(((μ−2)^2 )/2)} ...(ii) (i)/(ii): (√(4−λ^2 ))=(λ/(2−μ)) ⇒μ=2−(λ/( (√(4−λ^2 )))) put into (ii): ⇒4(λ−2+(λ/( (√(4−λ^2 )))))=(λ/( (√(4−λ^2 )))){(√(4−λ^2 ))−2−(λ^2 /(2(4−λ^2 )))} ⇒3λ+((6λ)/( (√(4−λ^2 ))))+(λ^3 /(2(4−λ^2 )(√(4−λ^2 ))))=8 ⇒λ=1.1672753 ⇒p=1.1672753 a ⇒c=0.8120142 a ....](https://www.tinkutara.com/question/Q59881.png)
$${method}\:\mathrm{1}: \\ $$$${eqn}.\:{of}\:{parabola}: \\ $$$${y}=\frac{\left({x}−\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{4}{a}}+{a} \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${A}\left({p},{c}\right) \\ $$$${B}\left({q},{d}\right) \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{c}={a}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{p}}{{a}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{c}=\frac{{a}}{\mathrm{2}}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} } \\ $$$${d}=\frac{\mathrm{1}}{\mathrm{4}{a}}\left({q}−\mathrm{2}{a}\right)^{\mathrm{2}} +{a}={a}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{q}}{{a}}−\mathrm{2}\right)^{\mathrm{2}} \right] \\ $$$$\Rightarrow{d}=\frac{{a}}{\mathrm{4}}\left[\mathrm{4}+\left(\mu−\mathrm{2}\right)^{\mathrm{2}} \right] \\ $$$${D}=\left({AB}\right)^{\mathrm{2}} =\left({p}−{q}\right)^{\mathrm{2}} +\left({c}−{d}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left[\left(\lambda−\mu\right)^{\mathrm{2}} +\left\{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{1}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}^{\mathrm{2}} \right]={a}^{\mathrm{2}} {F}\left(\lambda,\mu\right) \\ $$$${F}\left(\lambda,\mu\right)=\left(\lambda−\mu\right)^{\mathrm{2}} +\left\{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{1}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}^{\mathrm{2}} \\ $$$$\frac{\partial{F}}{\partial\lambda}=\mathrm{2}\left(\lambda−\mu\right)+\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{1}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}×\left(−\mathrm{2}\lambda\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\left(\lambda−\mu\right)\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }=\lambda\left\{\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{2}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}\right\}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial{F}}{\partial\mu}=−\mathrm{2}\left(\lambda−\mu\right)+\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{1}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}\left\{−\frac{\mathrm{2}\left(\mu−\mathrm{2}\right)}{\mathrm{4}}\right\}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\left(\lambda−\mu\right)=\left(\mathrm{2}−\mu\right)\left\{\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{2}−\frac{\left(\mu−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}\right\}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }=\frac{\lambda}{\mathrm{2}−\mu} \\ $$$$\Rightarrow\mu=\mathrm{2}−\frac{\lambda}{\:\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }} \\ $$$${put}\:{into}\:\left({ii}\right): \\ $$$$\Rightarrow\mathrm{4}\left(\lambda−\mathrm{2}+\frac{\lambda}{\:\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}\right)=\frac{\lambda}{\:\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}\left\{\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }−\mathrm{2}−\frac{\lambda^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{4}−\lambda^{\mathrm{2}} \right)}\right\}\: \\ $$$$\Rightarrow\mathrm{3}\lambda+\frac{\mathrm{6}\lambda}{\:\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}+\frac{\lambda^{\mathrm{3}} }{\mathrm{2}\left(\mathrm{4}−\lambda^{\mathrm{2}} \right)\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}=\mathrm{8} \\ $$$$\Rightarrow\lambda=\mathrm{1}.\mathrm{1672753} \\ $$$$\Rightarrow{p}=\mathrm{1}.\mathrm{1672753}\:{a} \\ $$$$\Rightarrow{c}=\mathrm{0}.\mathrm{8120142}\:{a} \\ $$$$…. \\ $$
Commented by ajfour last updated on 15/May/19
$$\mathrm{cant}\:\mathrm{we}\:\mathrm{do}\:\mathrm{it}\:\mathrm{with}\:\mathrm{one}\:\mathrm{variable}\:\mathrm{Sir}? \\ $$
Commented by mr W last updated on 16/May/19
$${yes}\:{sir}.\:\:{i}\:{also}\:{tried}\:{to}\:{do}\:{with}\:{only}\:{one}\: \\ $$$${variable}:\:{slope}\:{of}\:{tangents}\:{at}\:{A}\:{and}\:{B} \\ $$$${and}\:{without}\:{using}\:{calculus}. \\ $$$${both}\:{methods}\:{give}\:{the}\:{same}\:{result}. \\ $$
Answered by mr W last updated on 16/May/19
$${method}\:\mathrm{2}: \\ $$$${parabola}: \\ $$$${y}=\frac{\left({x}−\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{4}{a}}+{a} \\ $$$${ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${point}\:{A}\left({p},{c}\right) \\ $$$${point}\:{B}\left({q},{d}\right) \\ $$$$ \\ $$$${slope}\:{of}\:{tangent}\:{at}\:{A}\:{is}\:{m}: \\ $$$$\frac{\mathrm{2}{x}}{\mathrm{4}{a}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{{a}^{\mathrm{2}} }{y}'=\mathrm{0} \\ $$$$\frac{{p}}{\mathrm{4}}+{cm}=\mathrm{0} \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\Rightarrow{c}=\frac{{a}}{\:\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }} \\ $$$$\Rightarrow{p}=−\frac{\mathrm{4}{ma}}{\:\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }} \\ $$$$ \\ $$$${slope}\:{of}\:{tangent}\:{at}\:{B}\:{is}\:{also}\:{m}: \\ $$$${y}'=\frac{{x}−\mathrm{2}{a}}{\mathrm{2}{a}} \\ $$$${m}=\frac{{q}}{\mathrm{2}{a}}−\mathrm{1} \\ $$$$\Rightarrow{q}=\mathrm{2}{a}\left(\mathrm{1}+{m}\right) \\ $$$$\Rightarrow{d}=\left(\mathrm{1}+{m}^{\mathrm{2}} \right){a} \\ $$$$ \\ $$$${slope}\:{of}\:{AB}={m}_{\mathrm{1}} \\ $$$${m}_{\mathrm{1}} =\frac{{d}−{c}}{{q}−{p}}=\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right){a}−\frac{{a}}{\:\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }}}{\mathrm{2}{a}\left(\mathrm{1}+{m}\right)+\frac{\mathrm{4}{ma}}{\:\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }}}=\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{m}\right)\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }+\mathrm{4}{m}} \\ $$$${m}_{\mathrm{1}} ×{m}=−\mathrm{1} \\ $$$$\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{m}\right)\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }+\mathrm{4}{m}}×{m}=−\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{2}+\mathrm{3}{m}+{m}^{\mathrm{3}} \right)\sqrt{\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} }+\mathrm{3}{m}=\mathrm{0} \\ $$$$\Rightarrow{m}=−\mathrm{0}.\mathrm{3593765} \\ $$$$\Rightarrow{p}=\mathrm{1}.\mathrm{1672753}\:{a} \\ $$$$\Rightarrow{c}=\mathrm{0}.\mathrm{8120142}\:{a} \\ $$$$……. \\ $$
Commented by ajfour last updated on 16/May/19
$$\mathcal{VERY}\:\mathcal{NICE}\:{Sir}!\:{thanks}. \\ $$