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Question-65281




Question Number 65281 by mr W last updated on 27/Jul/19
Commented by mr W last updated on 27/Jul/19
segment of circle with size 2c×d. (d≤c)  find the inscribed ellipse with  maximum area.
segmentofcirclewithsize2c×d.(dc)findtheinscribedellipsewithmaximumarea.
Answered by mr W last updated on 28/Jul/19
R=radius of circle  d(2R−d)=c^2   ⇒R=((c^2 +d^2 )/(2d))  let h=R−d=((c^2 −d^2 )/(2d))  eqn. of circle:  x^2 +(y+h)^2 =R^2   x^2 +y^2 +2hy+h^2 −R^2 =0  ⇒x^2 +y^2 +2hy−c^2 =0    ...(i)  eqn. of ellipse:  (x^2 /a^2 )+(((y−b)^2 )/b^2 )=1  x^2 +(a^2 /b^2 )y^2 −((2a^2 )/b)y=0   ...(ii)  (i)−(ii):  (1−(a^2 /b^2 ))y^2 +2(h+(a^2 /b))y−c^2 =0  due to tangency there is only one root,  Δ=4(h+(a^2 /b))^2 +4c^2 (1−(a^2 /b^2 ))=0  ⇒(hb+a^2 )^2 +c^2 (b^2 −a^2 )=0  ...(iii)  ⇒(hab+a^3 )^2 +c^2 (a^2 b^2 −a^4 )=0  let P=ab  ⇒(hP+a^3 )^2 +c^2 (P^2 −a^4 )=0  for maximum ellipse area, (dP/da)=0  ⇒2(hP+a^3 )(3a^2 )+c^2 (−4a^3 )=0  ⇒3(hb+a^2 )−2c^2 =0  if h=0, i.e. d=c:  3a^2 −2c^2 =0⇒a=(((√6)c)/3)⇒b=(((√2)c)/3)  if h≠0:  ⇒b=(1/h)(((2c^2 )/3)−a^2 )  insert this into (iii):  (((2c^2 )/3)−a^2 +a^2 )^2 +c^2 (b^2 −a^2 )=0  ((4c^2 )/9)+(1/h^2 )(((2c^2 )/3)−a^2 )^2 −a^2 =0  ((4c^2 h^2 )/9)+(((2c^2 )/3)−a^2 )^2 −h^2 a^2 =0  a^4 −(((4c^2 )/3)+h^2 )a^2 +((4c^2 (c^2 +h^2 ))/9)=0  ⇒a^2 =(1/2)[((4c^2 )/3)+h^2 −(√((((4c^2 )/3)+h^2 )^2 −((16c^2 (c^2 +h^2 ))/9)))]  ⇒a^2 =(1/2)[((4c^2 )/3)+h^2 −(√((16c^4 +24c^2 h^2 +9h^4 −16c^4 −16c^2 h^2 )/9))]  ⇒a^2 =(1/2)(((4c^2 )/3)+h^2 −h(√(((8c^2 )/9)+h^2 )))  ⇒a=(√((1/2)(((4c^2 )/3)+h^2 −h(√(((8c^2 )/9)+h^2 )))))
R=radiusofcircled(2Rd)=c2R=c2+d22dleth=Rd=c2d22deqn.ofcircle:x2+(y+h)2=R2x2+y2+2hy+h2R2=0x2+y2+2hyc2=0(i)eqn.ofellipse:x2a2+(yb)2b2=1x2+a2b2y22a2by=0(ii)(i)(ii):(1a2b2)y2+2(h+a2b)yc2=0duetotangencythereisonlyoneroot,Δ=4(h+a2b)2+4c2(1a2b2)=0(hb+a2)2+c2(b2a2)=0(iii)(hab+a3)2+c2(a2b2a4)=0letP=ab(hP+a3)2+c2(P2a4)=0formaximumellipsearea,dPda=02(hP+a3)(3a2)+c2(4a3)=03(hb+a2)2c2=0ifh=0,i.e.d=c:3a22c2=0a=6c3b=2c3ifh0:b=1h(2c23a2)insertthisinto(iii):(2c23a2+a2)2+c2(b2a2)=04c29+1h2(2c23a2)2a2=04c2h29+(2c23a2)2h2a2=0a4(4c23+h2)a2+4c2(c2+h2)9=0a2=12[4c23+h2(4c23+h2)216c2(c2+h2)9]a2=12[4c23+h216c4+24c2h2+9h416c416c2h29]a2=12(4c23+h2h8c29+h2)a=12(4c23+h2h8c29+h2)
Commented by mr W last updated on 28/Jul/19

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