Question Number 164178 by mr W last updated on 15/Jan/22
Commented by mr W last updated on 15/Jan/22
$${two}\:{points}\:{P},\:{Q}\:{lie}\:{on}\:{the}\:{parabola} \\ $$$${with}\:{PQ}={b}.\:{find}\:{the}\:{locus}\:{of}\:{its} \\ $$$${midpoint}\:{M}.\:{find}\:{the}\:{lowest}\:{possible} \\ $$$${position}\:{of}\:{M}. \\ $$
Answered by mindispower last updated on 15/Jan/22
![P(a,a^2 );Q(b,b^2 ) M=(((a+b)/2),((a^2 +b^2 )/2)) PQ=l⇒(b−a)^2 +(b^2 −a^2 )^2 =l^2 (b−a)^2 (1+(b+a)^2 )=l^2 b−a=y b+a=z y^2 (1+z^2 )=l^2 a^2 +b^2 =((y^2 +z^2 )/2) we Want Min((1/8)(y^2 +z^2 )∵y^2 (1+z^2 )=l^2 f(x,y,γ)=(1/8)y^2 +z^2 −γ(y^2 (1+z^2 )−l^2 ) ∂_y f=0⇔(1/4)y−2yγ(1+z^2 )=0 y=0 or γ=(1/(8(1+z^2 ))) ∂zf=0⇔0⇒(1/4)z−2γy^2 z=0 z(1−8γy^2 )=0 y^2 (1+z^2 )=l^2 y=0⇒l=0,z=0 P=Q=O(0,0) l>0,z=0⇒y^2 =l^2 ,y(1−8γ)=0 y=b−a,0=a+b b=−a=(y/2)M=((a^2 +b^2 )/2)=(l^2 /4) γ=(1/(8(1+z^2 )))=(1/(8y^2 )) y^2 =1+z^2 y^4 =l^2 y^2 =l z^2 =l−1,l≥1 (l/4)≤(l^2 /4) l=0 min=0 l∈[0,1] min=(l^2 /4) l>1 min=(l/4) i did quick i will tchek it](https://www.tinkutara.com/question/Q164221.png)
$${P}\left({a},{a}^{\mathrm{2}} \right);{Q}\left({b},{b}^{\mathrm{2}} \right) \\ $$$${M}=\left(\frac{{a}+{b}}{\mathrm{2}},\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${PQ}={l}\Rightarrow\left({b}−{a}\right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} ={l}^{\mathrm{2}} \\ $$$$\left({b}−{a}\right)^{\mathrm{2}} \left(\mathrm{1}+\left({b}+{a}\right)^{\mathrm{2}} \right)={l}^{\mathrm{2}} \\ $$$${b}−{a}={y} \\ $$$${b}+{a}={z} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)={l}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${we}\:{Want}\:{Min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\because{y}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)={l}^{\mathrm{2}} \right. \\ $$$${f}\left({x},{y},\gamma\right)=\frac{\mathrm{1}}{\mathrm{8}}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\gamma\left({y}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)−\boldsymbol{{l}}^{\mathrm{2}} \right) \\ $$$$\partial_{{y}} {f}=\mathrm{0}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{4}}{y}−\mathrm{2}{y}\gamma\left(\mathrm{1}+{z}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${y}=\mathrm{0}\:{or}\:\gamma=\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)} \\ $$$$\partial{zf}=\mathrm{0}\Leftrightarrow\mathrm{0}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}{z}−\mathrm{2}\gamma{y}^{\mathrm{2}} {z}=\mathrm{0} \\ $$$${z}\left(\mathrm{1}−\mathrm{8}\gamma{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)={l}^{\mathrm{2}} \\ $$$${y}=\mathrm{0}\Rightarrow{l}=\mathrm{0},{z}=\mathrm{0}\:{P}={Q}={O}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${l}>\mathrm{0},{z}=\mathrm{0}\Rightarrow{y}^{\mathrm{2}} ={l}^{\mathrm{2}} ,{y}\left(\mathrm{1}−\mathrm{8}\gamma\right)=\mathrm{0} \\ $$$${y}={b}−{a},\mathrm{0}={a}+{b} \\ $$$${b}=−{a}=\frac{{y}}{\mathrm{2}}{M}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\gamma=\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} =\mathrm{1}+{z}^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} ={l}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} ={l} \\ $$$${z}^{\mathrm{2}} ={l}−\mathrm{1},{l}\geqslant\mathrm{1} \\ $$$$\frac{{l}}{\mathrm{4}}\leqslant\frac{{l}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${l}=\mathrm{0}\:{min}=\mathrm{0} \\ $$$${l}\in\left[\mathrm{0},\mathrm{1}\right]\:{min}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${l}>\mathrm{1}\:{min}=\frac{{l}}{\mathrm{4}}\:{i}\:{did}\:{quick}\:{i}\:{will}\:{tchek}\:{it} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 15/Jan/22
$${thanks}\:{sir}! \\ $$
Commented by mindispower last updated on 15/Jan/22
$${pleasur}\:{sir}\:{nice}\:{day} \\ $$
Answered by mr W last updated on 15/Jan/22
![parabola y=(x^2 /a) with a=1 here say P(p,(p^2 /a)), Q(q,(q^2 /a)) say M(u,v) u=((p+q)/2) v=((p^2 +q^2 )/(2a)) ⇒p+q=2u ⇒p^2 +q^2 =2av (p+q)^2 =p^2 +q^2 +2pq (p−q)^2 =p^2 +q^2 −2pq ⇒(p−q)^2 +(p+q)^2 =2(p^2 +q^2 ) ⇒(p−q)^2 =2(2av)−(2u)^2 =4(av−u^2 ) PQ^2 =(p−q)^2 +((p^2 /a)−(q^2 /a))^2 =b^2 (p−q)^2 [1+(((p+q)^2 )/a^2 )]=b^2 4(av−u^2 )[1+(((2u)^2 )/a^2 )]=b^2 4(av−u^2 )(1+((4u^2 )/a^2 ))=b^2 v=(u^2 /a)+((ab^2 )/(4(a^2 +4u^2 ))) or locus of point M is y=(x^2 /a)+((ab^2 )/(4(a^2 +4x^2 ))) y=((4x^2 +a^2 )/(4a))+((ab^2 )/(4(a^2 +4x^2 )))−(a/4) ≥2(√(((4x^2 +a^2 )/(4a))×((ab^2 )/(4(a^2 +4x^2 )))))−(a/4) =(b/2)−(a/4) i.e. y_(min) =(b/2)−(a/4) when ((4x^2 +a^2 )/(4a))=((ab^2 )/(4(a^2 +4x^2 ))) 4x^2 +a^2 =ab x=±((√(a(b−a)))/2) that means the lowest position of M is (±((√(a(b−a)))/2), (b/2)−(a/4))](https://www.tinkutara.com/question/Q164280.png)
$${parabola}\:{y}=\frac{{x}^{\mathrm{2}} }{{a}}\:{with}\:{a}=\mathrm{1}\:{here} \\ $$$${say}\:{P}\left({p},\frac{{p}^{\mathrm{2}} }{{a}}\right),\:{Q}\left({q},\frac{{q}^{\mathrm{2}} }{{a}}\right) \\ $$$${say}\:{M}\left({u},{v}\right) \\ $$$${u}=\frac{{p}+{q}}{\mathrm{2}} \\ $$$${v}=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$$\Rightarrow{p}+{q}=\mathrm{2}{u} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{2}{av} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{2}{pq} \\ $$$$\left({p}−{q}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pq} \\ $$$$\Rightarrow\left({p}−{q}\right)^{\mathrm{2}} +\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left({p}−{q}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}{av}\right)−\left(\mathrm{2}{u}\right)^{\mathrm{2}} =\mathrm{4}\left({av}−{u}^{\mathrm{2}} \right) \\ $$$${PQ}^{\mathrm{2}} =\left({p}−{q}\right)^{\mathrm{2}} +\left(\frac{{p}^{\mathrm{2}} }{{a}}−\frac{{q}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({p}−{q}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\left({p}+{q}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \\ $$$$\mathrm{4}\left({av}−{u}^{\mathrm{2}} \right)\left[\mathrm{1}+\frac{\left(\mathrm{2}{u}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \\ $$$$\mathrm{4}\left({av}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{4}{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)={b}^{\mathrm{2}} \\ $$$${v}=\frac{{u}^{\mathrm{2}} }{{a}}+\frac{{ab}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{4}{u}^{\mathrm{2}} \right)} \\ $$$${or}\:{locus}\:{of}\:{point}\:{M}\:{is} \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{{a}}+\frac{{ab}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${y}=\frac{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mathrm{4}{a}}+\frac{{ab}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \right)}−\frac{{a}}{\mathrm{4}} \\ $$$$\:\:\geqslant\mathrm{2}\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mathrm{4}{a}}×\frac{{ab}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \right)}}−\frac{{a}}{\mathrm{4}} \\ $$$$\:\:=\frac{{b}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}} \\ $$$${i}.{e}.\:\:{y}_{{min}} =\frac{{b}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}}\:{when} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mathrm{4}{a}}=\frac{{ab}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \right)} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} ={ab} \\ $$$${x}=\pm\frac{\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$${that}\:{means}\:{the}\:{lowest}\:{position}\:{of}\:{M} \\ $$$${is}\:\left(\pm\frac{\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}},\:\frac{{b}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}}\right) \\ $$
Commented by mr W last updated on 15/Jan/22
Commented by Tawa11 last updated on 15/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$