Question Number 169669 by ajfour last updated on 05/May/22
Answered by mr W last updated on 06/May/22
Commented by mr W last updated on 06/May/22
![P(p,q) ((a−q)/p)=(a/b) ⇒q=a(1−(p/b)) eqn. of BM: (x/p)+(y/a)=1 eqn. of AN: (x/b)+(y/q)=1 ax_Q +py_Q =ap qx_Q +by_Q =bq ⇒x_Q =((bp(a−q))/(ab−pq)) ⇒y_Q =((aq(b−p))/(ab−pq)) PQ^2 =[p−((bp(a−q))/(ab−pq))]^2 +[q−((aq(b−p))/(ab−pq))]^2 PQ^2 =(([(b−p)^2 +(a−q)^2 ]p^2 q^2 )/((ab−pq)^2 )) PQ^2 =(((a^2 +b^2 +p^2 +q^2 −2bp−2aq)p^2 q^2 )/((ab−pq)^2 )) let λ=(a/b), ξ=(p/b), μ=(q/a)=1−ξ ((PQ^2 )/b^2 )=(((λ^2 +1+ξ^2 +λ^2 μ^2 −2ξ−2λ^2 μ)ξ^2 μ^2 )/((1−ξμ)^2 )) ((PQ^2 )/b^2 )=(((λ^2 +1+ξ^2 +λ^2 (1−ξ)^2 −2ξ−2λ^2 (1−ξ))ξ^2 (1−ξ)^2 )/((1−ξ(1−ξ))^2 )) ((PQ^2 )/b^2 )=(([(λ^2 +1)ξ^2 −2ξ+1]ξ^2 (1−ξ)^2 )/((ξ^2 −ξ+1)^2 )) ((PQ)/b)=Φ=((ξ(1−ξ)(√((λ^2 +1)ξ^2 −2ξ+1)))/(ξ^2 −ξ+1)) such that PQ is maximum, (dΦ/dξ)=0 example: λ=(a/b)=0.5 ⇒ξ=0.387, Φ=0.1999 ⇒PQ_(max) =0.1999b example: λ=(a/b)=1 ⇒ξ=0.5, Φ=0.2357 ⇒PQ_(max) =0.2357b](Q169684.png)
$${P}\left({p},{q}\right) \\ $$$$\frac{{a}−{q}}{{p}}=\frac{{a}}{{b}}\:\Rightarrow{q}={a}\left(\mathrm{1}−\frac{{p}}{{b}}\right) \\ $$$${eqn}.\:{of}\:{BM}: \\ $$$$\frac{{x}}{{p}}+\frac{{y}}{{a}}=\mathrm{1} \\ $$$${eqn}.\:{of}\:{AN}: \\ $$$$\frac{{x}}{{b}}+\frac{{y}}{{q}}=\mathrm{1} \\ $$$$ \\ $$$${ax}_{{Q}} +{py}_{{Q}} ={ap} \\ $$$${qx}_{{Q}} +{by}_{{Q}} ={bq} \\ $$$$\Rightarrow{x}_{{Q}} =\frac{{bp}\left({a}−{q}\right)}{{ab}−{pq}} \\ $$$$\Rightarrow{y}_{{Q}} =\frac{{aq}\left({b}−{p}\right)}{{ab}−{pq}} \\ $$$${PQ}^{\mathrm{2}} =\left[{p}−\frac{{bp}\left({a}−{q}\right)}{{ab}−{pq}}\right]^{\mathrm{2}} +\left[{q}−\frac{{aq}\left({b}−{p}\right)}{{ab}−{pq}}\right]^{\mathrm{2}} \\ $$$${PQ}^{\mathrm{2}} =\frac{\left[\left({b}−{p}\right)^{\mathrm{2}} +\left({a}−{q}\right)^{\mathrm{2}} \right]{p}^{\mathrm{2}} {q}^{\mathrm{2}} }{\left({ab}−{pq}\right)^{\mathrm{2}} } \\ $$$${PQ}^{\mathrm{2}} =\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{bp}−\mathrm{2}{aq}\right){p}^{\mathrm{2}} {q}^{\mathrm{2}} }{\left({ab}−{pq}\right)^{\mathrm{2}} } \\ $$$${let}\:\lambda=\frac{{a}}{{b}},\:\xi=\frac{{p}}{{b}},\:\mu=\frac{{q}}{{a}}=\mathrm{1}−\xi \\ $$$$\frac{{PQ}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\left(\lambda^{\mathrm{2}} +\mathrm{1}+\xi^{\mathrm{2}} +\lambda^{\mathrm{2}} \mu^{\mathrm{2}} −\mathrm{2}\xi−\mathrm{2}\lambda^{\mathrm{2}} \mu\right)\xi^{\mathrm{2}} \mu^{\mathrm{2}} }{\left(\mathrm{1}−\xi\mu\right)^{\mathrm{2}} } \\ $$$$\frac{{PQ}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\left(\lambda^{\mathrm{2}} +\mathrm{1}+\xi^{\mathrm{2}} +\lambda^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} −\mathrm{2}\xi−\mathrm{2}\lambda^{\mathrm{2}} \left(\mathrm{1}−\xi\right)\right)\xi^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} }{\left(\mathrm{1}−\xi\left(\mathrm{1}−\xi\right)\right)^{\mathrm{2}} } \\ $$$$\frac{{PQ}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\left[\left(\lambda^{\mathrm{2}} +\mathrm{1}\right)\xi^{\mathrm{2}} −\mathrm{2}\xi+\mathrm{1}\right]\xi^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} }{\left(\xi^{\mathrm{2}} −\xi+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{PQ}}{{b}}=\Phi=\frac{\xi\left(\mathrm{1}−\xi\right)\sqrt{\left(\lambda^{\mathrm{2}} +\mathrm{1}\right)\xi^{\mathrm{2}} −\mathrm{2}\xi+\mathrm{1}}}{\xi^{\mathrm{2}} −\xi+\mathrm{1}} \\ $$$${such}\:{that}\:{PQ}\:{is}\:{maximum}, \\ $$$$\frac{{d}\Phi}{{d}\xi}=\mathrm{0} \\ $$$$ \\ $$$${example}:\:\lambda=\frac{{a}}{{b}}=\mathrm{0}.\mathrm{5} \\ $$$$\Rightarrow\xi=\mathrm{0}.\mathrm{387},\:\Phi=\mathrm{0}.\mathrm{1999}\: \\ $$$$\Rightarrow{PQ}_{{max}} =\mathrm{0}.\mathrm{1999}{b} \\ $$$${example}:\:\lambda=\frac{{a}}{{b}}=\mathrm{1} \\ $$$$\Rightarrow\xi=\mathrm{0}.\mathrm{5},\:\Phi=\mathrm{0}.\mathrm{2357}\: \\ $$$$\Rightarrow{PQ}_{{max}} =\mathrm{0}.\mathrm{2357}{b} \\ $$
Commented by mr W last updated on 06/May/22
Commented by ajfour last updated on 06/May/22
$${thanks}\:{sir},\:{i}\:{had}\:{expected}\: \\ $$$${the}\:{question}\:{to}\:{be}\:{an}\:{easy}\:{one}. \\ $$
Commented by Tawa11 last updated on 06/May/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$