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Question Number 112356 by ajfour last updated on 07/Sep/20
Commented by ajfour last updated on 07/Sep/20
Find minimum length of PQ in terms of a and b.
$${Find}\:{minimum}\:{length}\:{of}\:{PQ}\:{in} \\ $$$${terms}\:{of}\:{a}\:{and}\:{b}.\:\:\:\:\:\:\: \\ $$
Answered by bemath last updated on 07/Sep/20
let Q(x_1 ,x_1 ^2 +b) tangent line at Q⇒((y+x_1 ^2 +b)/2)=x_1 x +b y=2x_1 x+b−x_1 ^2 , with gradient m_1 = 2x_1 let P(y_2 ^2 +a , y_2 ) , tangent at P parallel to tangent line at P ((x+y_2 ^2 +a)/2) = y_2 y +a y = ((x+y_2 ^2 −a)/y_2 ) with gradient m_2 =(1/y_2 ) so 2x_1 = (1/y_2 ) or y_2 =(1/(2x_1 )) therefore { ((Q(x_1 ,x_1 ^2 +b))),((P((1/(4x_1 ^2 ))+a , (1/(2x_1 ^2 ))))) :} then PQ=(√((x_1 −a−(1/(4x_1 ^2 )))^2 +(x_1 ^2 +b−(1/(2x_1 ^2 )))^2 ))
$$\mathrm{let}\:\mathrm{Q}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}\right) \\ $$$$\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{Q}\Rightarrow\frac{\mathrm{y}+\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}}{\mathrm{2}}=\mathrm{x}_{\mathrm{1}} \mathrm{x}\:+\mathrm{b} \\ $$$$\mathrm{y}=\mathrm{2x}_{\mathrm{1}} \mathrm{x}+\mathrm{b}−\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \:,\:\mathrm{with}\:\mathrm{gradient} \\ $$$$\mathrm{m}_{\mathrm{1}} =\:\mathrm{2x}_{\mathrm{1}} \\ $$$$\mathrm{let}\:\mathrm{P}\left(\mathrm{y}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{a}\:,\:\mathrm{y}_{\mathrm{2}} \right)\:,\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{P}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{P} \\ $$$$\frac{\mathrm{x}+\mathrm{y}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{a}}{\mathrm{2}}\:=\:\mathrm{y}_{\mathrm{2}} \mathrm{y}\:+\mathrm{a} \\ $$$$\mathrm{y}\:=\:\frac{\mathrm{x}+\mathrm{y}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{a}}{\mathrm{y}_{\mathrm{2}} }\:\mathrm{with}\:\mathrm{gradient}\:\mathrm{m}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{y}_{\mathrm{2}} } \\ $$$$\mathrm{so}\:\mathrm{2x}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{y}_{\mathrm{2}} }\:\mathrm{or}\:\mathrm{y}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2x}_{\mathrm{1}} } \\ $$$$\mathrm{therefore}\:\begin{cases}{\mathrm{Q}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}\right)}\\{\mathrm{P}\left(\frac{\mathrm{1}}{\mathrm{4x}_{\mathrm{1}} ^{\mathrm{2}} }+\mathrm{a}\:,\:\frac{\mathrm{1}}{\mathrm{2x}_{\mathrm{1}} ^{\mathrm{2}} }\right)}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{PQ}=\sqrt{\left(\mathrm{x}_{\mathrm{1}} −\mathrm{a}−\frac{\mathrm{1}}{\mathrm{4x}_{\mathrm{1}} ^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}−\frac{\mathrm{1}}{\mathrm{2x}_{\mathrm{1}} ^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 07/Sep/20
Sir please find x_1 such that PQ be min.
$${Sir}\:{please}\:{find}\:{x}_{\mathrm{1}} \:{such}\:{that}\:{PQ}\:{be}\:{min}. \\ $$
Answered by mr W last updated on 07/Sep/20
P(a+p^2 ,p) Q(q,b+q^2 ) (dy/dx)∣_Q =2q (dy/dx)∣_P =(1/(dx/dy))=(1/(2p)) ⇒2q=(1/(2p)) ⇒pq=(1/4) ((b+q^2 −p)/(q−a−p^2 ))=−2p b+q^2 −p=−2pq+2ap+2p^3 b+(1/(16p^2 ))−p=−(1/2)+2ap+2p^3 32p^5 +16(2a+1)p^3 −8(1+2b)p^2 −1=0 λ=2p ⇒λ^5 +2(1+2a)λ^3 −2(1+2b)λ^2 −1=0
$${P}\left({a}+{p}^{\mathrm{2}} ,{p}\right) \\ $$$${Q}\left({q},{b}+{q}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{dx}}\mid_{{Q}} =\mathrm{2}{q} \\ $$$$\frac{{dy}}{{dx}}\mid_{{P}} =\frac{\mathrm{1}}{\frac{{dx}}{{dy}}}=\frac{\mathrm{1}}{\mathrm{2}{p}} \\ $$$$\Rightarrow\mathrm{2}{q}=\frac{\mathrm{1}}{\mathrm{2}{p}} \\ $$$$\Rightarrow{pq}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{{b}+{q}^{\mathrm{2}} −{p}}{{q}−{a}−{p}^{\mathrm{2}} }=−\mathrm{2}{p} \\ $$$${b}+{q}^{\mathrm{2}} −{p}=−\mathrm{2}{pq}+\mathrm{2}{ap}+\mathrm{2}{p}^{\mathrm{3}} \\ $$$${b}+\frac{\mathrm{1}}{\mathrm{16}{p}^{\mathrm{2}} }−{p}=−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{ap}+\mathrm{2}{p}^{\mathrm{3}} \\ $$$$\mathrm{32}{p}^{\mathrm{5}} +\mathrm{16}\left(\mathrm{2}{a}+\mathrm{1}\right){p}^{\mathrm{3}} −\mathrm{8}\left(\mathrm{1}+\mathrm{2}{b}\right){p}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\lambda=\mathrm{2}{p} \\ $$$$\Rightarrow\lambda^{\mathrm{5}} +\mathrm{2}\left(\mathrm{1}+\mathrm{2}{a}\right)\lambda^{\mathrm{3}} −\mathrm{2}\left(\mathrm{1}+\mathrm{2}{b}\right)\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$
Commented by mr W last updated on 07/Sep/20
Commented by ajfour last updated on 07/Sep/20
Thank you Sir ; very well presented!
$${Thank}\:{you}\:{Sir}\:;\:{very}\:{well}\:{presented}! \\ $$
Answered by ajfour last updated on 07/Sep/20
P (p^2 +a, p) ; Q(q, q^2 +b) eqn. of normal at P y−p=−2p(x−p^2 −a) eqn. of normal at Q y−q^2 −b=−(1/(2q))(x−q) −−−−−−−−−−−−−−−−− pq=(1/4) & p+2p^3 +2ap= q^2 +b+(1/2) −−−−−−−−−−−−−−−−− ⇒ p+2p(p^2 +a)=q^2 +b+(1/2) L_(PQ) ^2 =(p^2 −q+a)^2 +(q^2 −p+b)^2 = (p^2 +a−q)^2 +{2p(p^2 +a)−(1/2)}^2 = ((1/(4p^2 ))+1){2p(p^2 +a)−(1/2)}^2 (L_(PQ) )_(min) =(√(1+4p^2 ))(p^2 +a−(1/(4p))) −−−−−−−−−−−−−−−−
$${P}\:\left({p}^{\mathrm{2}} +{a},\:{p}\right)\:\:\:;\:\:{Q}\left({q},\:{q}^{\mathrm{2}} +{b}\right) \\ $$$${eqn}.\:{of}\:{normal}\:{at}\:{P} \\ $$$${y}−{p}=−\mathrm{2}{p}\left({x}−{p}^{\mathrm{2}} −{a}\right) \\ $$$${eqn}.\:{of}\:{normal}\:{at}\:{Q} \\ $$$${y}−{q}^{\mathrm{2}} −{b}=−\frac{\mathrm{1}}{\mathrm{2}{q}}\left({x}−{q}\right) \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$\:{pq}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\&\:\:\:{p}+\mathrm{2}{p}^{\mathrm{3}} +\mathrm{2}{ap}=\:{q}^{\mathrm{2}} +{b}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$\Rightarrow\:\:{p}+\mathrm{2}{p}\left({p}^{\mathrm{2}} +{a}\right)={q}^{\mathrm{2}} +{b}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${L}_{{PQ}} ^{\mathrm{2}} =\left({p}^{\mathrm{2}} −{q}+{a}\right)^{\mathrm{2}} +\left({q}^{\mathrm{2}} −{p}+{b}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\:\left({p}^{\mathrm{2}} +{a}−{q}\right)^{\mathrm{2}} +\left\{\mathrm{2}{p}\left({p}^{\mathrm{2}} +{a}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\:\left(\frac{\mathrm{1}}{\mathrm{4}{p}^{\mathrm{2}} }+\mathrm{1}\right)\left\{\mathrm{2}{p}\left({p}^{\mathrm{2}} +{a}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\left({L}_{{PQ}} \right)_{{min}} \:=\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }\left({p}^{\mathrm{2}} +{a}−\frac{\mathrm{1}}{\mathrm{4}{p}}\right) \\ $$$$\:\:\:−−−−−−−−−−−−−−−− \\ $$

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