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Question-54447




Question Number 54447 by ajfour last updated on 03/Feb/19
Commented by ajfour last updated on 14/Feb/19
Find coordinates of A and C.
FindcoordinatesofAandC.
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
solve  x^2 =ry and (x−r)^2 +(y−r)^2 =r^2   (x−r)^2 +((x^2 /r)−r)^2 =r^2   x^2 −2xr+r^2 +(x^4 /r^2 )−2x^2 +r^2 =r^2   (x^4 /r^2 )−x^2 −2xr+r^2 =0  x^4 −r^2 x^2 −2xr^3 +r^4 =0  on solving  we get four value of x  so corresponding four value of y  thus we can get co iridinate of  point A and C  ok le t try to solve...  let (x/r)=k  (x^4 /r^4 )−(x^2 /r^2 )−((2x)/r)+1=0  k^4 −k^2 −2k+1=0  f(k)=k^4 −k^2 −2k+1=k^4 −k^2 −2k−1+2=k^4 +2−(k+1)^2   f(0)>0  f(1)<0  f(0.5)=(1/(16))−(1/4)−(2/2)+1<0  f(0.4)=0.0256+2−1.96>0  so value of [0.5 >k>0.4]  thus    0.5>(x/r)>0.4  so   0.5r>x>0.4r  thus  x=(0.4+ε)r   ε=is a small number...
solvex2=ryand(xr)2+(yr)2=r2(xr)2+(x2rr)2=r2x22xr+r2+x4r22x2+r2=r2x4r2x22xr+r2=0x4r2x22xr3+r4=0onsolvingwegetfourvalueofxsocorrespondingfourvalueofythuswecangetcoiridinateofpointAandCoklettrytosolveletxr=kx4r4x2r22xr+1=0k4k22k+1=0f(k)=k4k22k+1=k4k22k1+2=k4+2(k+1)2f(0)>0f(1)<0f(0.5)=1161422+1<0f(0.4)=0.0256+21.96>0sovalueof[0.5>k>0.4]thus0.5>xr>0.4so0.5r>x>0.4rthusx=(0.4+ϵ)rϵ=isasmallnumber
Commented by ajfour last updated on 03/Feb/19
why not solve it, Sir ?
whynotsolveit,Sir?
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
i can not solve without numirical value..  if numirical value present i can use calculus..  but general eqn...i can not solve...
icannotsolvewithoutnumiricalvalue..ifnumiricalvaluepresenticanusecalculus..butgeneraleqnicannotsolve
Commented by ajfour last updated on 03/Feb/19
it can be converted to one like  numerical value, but still let us not  use calculator.
itcanbeconvertedtoonelikenumericalvalue,butstillletusnotusecalculator.
Commented by ajfour last updated on 03/Feb/19
thanks for trying, Sir.
thanksfortrying,Sir.
Answered by ajfour last updated on 04/Feb/19
(x−r)^2 +(y−r)^2 =r^2     (circle eq.)  y=x^2 /r             (parabola eq.)  for x= x_A , x_C   x^2 −2rx+(x^4 /r^2 )−2x^2 +r^2 =0         (x^4 /r^2 )−x^2 −2rx+r^2 =0  if   x/r =t   ⇒   t^4 −t^2 −2t+1=0      ....(i)  let us assume equivalently       (t^2 −pt+q)(t^2 +pt+(1/q))=0  ⇒ t^4 +(q+(1/q)−p^2 )t^2 +(pq−(p/q))t+1=0  comparing with (i)      q+(1/q)= p^2 −1   and  q−(1/q)=−(2/p)  ⇒  (p^2 −1)^2 −(4/p^2 ) = 4  let us call  p^2 =s  ⇒  s(s−1)^2 −4s−4=0  let s−1= u  ⇒  (u+1)u^2 −4(u+2)=0  ⇒   u^3 +u^2 −4u−8=0  let  u=z−1/3  ⇒  z^3 −z^2 _− +(z/3)−(1/(27))+z^2 _− −((2z)/3)+(1/9)            −4z+(4/3)−8=0  or   z^3 −((13)/3)z−((178)/(27))=0      z= (((89)/(27))+(√(((89^2 )/(27^2 ))−((13^3 )/(27^2 )))) )^(1/3)                    +(((89)/(27))−(√((((89)/(27^2 ))−((13^3 )/(27^2 )))) )^(1/3)      =(1/3){(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3)   now   p^2 =s=u+1 = z+(2/3)  since t_A >0 and t_C  >0  let′s first choose  (t^2 −pt+q)=0  with p_1 >0   p_1 =(√((2+(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3) )/3))      ≈ 1.81226  (D/4)= (p^2 /4)−q    as  q= (((p^2 −1))/2)−(1/p)  (D/4)=(1/2)+(1/p)−(p^2 /4) ≈ 0.23073  if it is positive then  __________________________  with   p_1 =(√((2+(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3) )/3))    x_A =r( (p_1 /2)−(√((1/2)+(1/p_1 )−(p_1 ^2 /4))))  ⇒  x_A ≈ 0.4258 r   ,   y_A =x_A ^2 /r        x_C =r( (p_1 /2)+(√((1/2)+(1/p_1 )−(p_1 ^2 /4))))  ⇒  x_C  ≈ 1.3865 r   ,    y_C =x_C ^2 /r   __________________________■
(xr)2+(yr)2=r2(circleeq.)y=x2/r(parabolaeq.)forx=xA,xCx22rx+x4r22x2+r2=0x4r2x22rx+r2=0ifx/r=tt4t22t+1=0.(i)letusassumeequivalently(t2pt+q)(t2+pt+1q)=0t4+(q+1qp2)t2+(pqpq)t+1=0comparingwith(i)q+1q=p21andq1q=2p(p21)24p2=4letuscallp2=ss(s1)24s4=0lets1=u(u+1)u24(u+2)=0u3+u24u8=0letu=z1/3z3z2+z3127+z22z3+194z+438=0orz3133z17827=0z=(8927+892272133272)1/3+(8927(89272133272)1/3=13{(89+6159)1/3+(896159)1/3nowp2=s=u+1=z+23sincetA>0andtC>0letsfirstchoose(t2pt+q)=0withp1>0p1=2+(89+6159)1/3+(896159)1/331.81226D4=p24qasq=(p21)21pD4=12+1pp240.23073ifitispositivethen__________________________withp1=2+(89+6159)1/3+(896159)1/33xA=r(p1212+1p1p124)xA0.4258r,yA=xA2/rxC=r(p12+12+1p1p124)xC1.3865r,yC=xC2/r__________________________◼

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