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Question Number 58132 by ajfour last updated on 18/Apr/19

Commented by ajfour last updated on 18/Apr/19

Find coordinates of P and Q, in terms of a,b,c, and R.

FindcoordinatesofPandQ,intermsofa,b,c,andR.

Answered by tanmay last updated on 18/Apr/19

parabola y=αx^2 +βx+γ 0=αa^2 +βa+γ 0=αb^2 +β(−b)+γ −c=α×0^2 +β×0^2 +γ [ γ=−c] y=αx^2 +βx−c αa^2 +βa=αb^2 −βb α(a+b)(a−b)+β(a+b)=0 α(a−b)+β=0 β=α(b−a) 0=αa^2 +βa+γ 0=αa^2 +αa(b−a)−c c=α(a^2 +ab−a^2 ) α=(c/(ab)) β=(c/(ab))(b−a) parabola eqn y=((c/(ab)))x^2 +(c/(ab))(b−a)x−c←chek it again R+c=a+b radius ofcircle=a+b−c eqn of circle x^2 +y^2 =(a+b−c)^2 ←check it pls check eqn of circle and parabola... if correct...then to solve to find P and Q

parabolay=αx2+βx+γ0=αa2+βa+γ0=αb2+β(−b)+γ−c=α×02+β×02+γ[γ=−c]y=αx2+βx−cαa2+βa=αb2−βbα(a+b)(a−b)+β(a+b)=0α(a−b)+β=0β=α(b−a)0=αa2+βa+γ0=αa2+αa(b−a)−cc=α(a2+ab−a2)α=cabβ=cab(b−a)parabolaeqny=(cab)x2+cab(b−a)x−c←chekitagainR+c=a+bradiusofcircle=a+b−ceqnofcirclex2+y2=(a+b−c)2←checkitplscheckeqnofcircleandparabola...ifcorrect...thentosolvetofindPandQ

Commented by 121194 last updated on 18/Apr/19

x^2 +y^2 =R^2 (no relation)

x2+y2=R2(norelation)

Commented by tanmay last updated on 18/Apr/19

let radius of cirle be l eqn x^2 +y^2 =l^2 from figure R(0,l) so l+c=a+b l=a+b−c hence circle is x^2 +y^2 =(a+b−c)^2

letradiusofcirlebeleqnx2+y2=l2fromfigureR(0,l)sol+c=a+bl=a+b−chencecircleisx2+y2=(a+b−c)2

Commented by ajfour last updated on 18/Apr/19

never mind Sir, it would of course land on solving a general biquadratric, i understand..have gotten bit busy these days, shall post a good question soon.

nevermindSir,itwouldofcourselandonsolvingageneralbiquadratric,iunderstand..havegottenbitbusythesedays,shallpostagoodquestionsoon.

Answered by mr W last updated on 18/Apr/19

eqn. of parabola: y=α(x−((a+b)/2))^2 +β 0=α(a−((a+b)/2))^2 +β ⇒0=α(((a−b)/2))^2 +β ..(i) c=α(0−((a+b)/2))^2 +β ⇒c=α(((a+b)/2))^2 +β ..(ii) (ii)−(i): αab=c ⇒α=(c/(ab)) ⇒β=−((c(a−b)^2 )/(4ab)) let P,Q=(u,v) v=α(u−((a+b)/2))^2 +β ...(iv) u^2 +v^2 =R^2 u^2 +[α(u−((a+b)/2))^2 +β]^2 =R^2 ⇒u^2 +[(c/(ab))(u−((a+b)/2))^2 −((c(a−b)^2 )/(4ab))]^2 =R^2 ⇒u^2 +((c/(ab)))^2 (u−a)^2 (u−b)^2 =R^2 ...(iii) from (iii) u=f(a,b,c,R)=....... (iii) may have no solution, one solution, two, three or four solutions.

eqn.ofparabola:y=α(x−a+b2)2+β0=α(a−a+b2)2+β⇒0=α(a−b2)2+β..(i)c=α(0−a+b2)2+β⇒c=α(a+b2)2+β..(ii)(ii)−(i):αab=c⇒α=cab⇒β=−c(a−b)24abletP,Q=(u,v)v=α(u−a+b2)2+β...(iv)u2+v2=R2u2+[α(u−a+b2)2+β]2=R2⇒u2+[cab(u−a+b2)2−c(a−b)24ab]2=R2⇒u2+(cab)2(u−a)2(u−b)2=R2...(iii)from(iii)u=f(a,b,c,R)=.......(iii)mayhavenosolution,onesolution,two,threeorfoursolutions.

Commented by ajfour last updated on 20/Apr/19

thanks sir, but i think we can arrive at the final eq. even in the following manner: y=A(x−a)(x−b) & since −c=Aab ⇒ A=−(c/(ab)) so y=−(c/(ab))(x−a)(x−b) and circle eq. is x^2 +y^2 =R^2 hence for x coordinates of intersection points x^2 +((c/(ab)))^2 (x−a)^2 (x−b)^2 =R^2 .

thankssir,butithinkwecanarriveatthefinaleq.eveninthefollowingmanner:y=A(x−a)(x−b)&since−c=Aab⇒A=−cabsoy=−cab(x−a)(x−b)andcircleeq.isx2+y2=R2henceforxcoordinatesofintersectionpointsx2+(cab)2(x−a)2(x−b)2=R2.

Commented by mr W last updated on 20/Apr/19

very smart sir!

verysmartsir!

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