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Question Number 131642 by liberty last updated on 07/Feb/21

What is the maximum area   of ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 which touches  the line y = 3x+2.

Whatisthemaximumareaofellipsex2a2+y2b2=1whichtouchestheliney=3x+2.

Answered by benjo_mathlover last updated on 07/Feb/21

Area ellipse = πab  touches the line y=3x+2  by tangency ⇒y=mx+(√(a^2 m^2 +b^2 ))  where  { ((m=3)),(((√(9a^2 +b^2 )) = 2)) :}  or b=(√(4−9a^2 ))  then A=f(a)=πa(√(4−9a^2 ))  A^2  = π^2 a^2 (4−9a^2 )  9π^2 (a^2 )^2 −4π^2 a^2 +A^2 =0  Δ = 16π^4 −4(9π^2 )A^2 =0  ⇔ 4π^2 =9A^2  ⇒A=((2π)/3) ←max area

Areaellipse=πabtouchestheliney=3x+2bytangencyy=mx+a2m2+b2where{m=39a2+b2=2orb=49a2thenA=f(a)=πa49a2A2=π2a2(49a2)9π2(a2)24π2a2+A2=0Δ=16π44(9π2)A2=04π2=9A2A=2π3maxarea

Answered by mr W last updated on 07/Feb/21

area of ellipse is  A=πab  since it touches the line 3x−y+2=0,  (3a)^2 +(−1×b)^2 =2^2   ⇒b^2 =4−9a^2     A=π(√(a^2 b^2 ))=π(√(a^2 (4−9a^2 )))=(π/3)(√(9a^2 (4−9a^2 )))  ≤(π/3)×((9a^2 +(4−9a^2 ))/2)=((2π)/3)  i.e. A_(max) =((2π)/3)  when 9a^2 =4−9a^2  ⇒a=((√2)/3) ⇒b=3a=(√2)

areaofellipseisA=πabsinceittouchestheline3xy+2=0,(3a)2+(1×b)2=22b2=49a2A=πa2b2=πa2(49a2)=π39a2(49a2)π3×9a2+(49a2)2=2π3i.e.Amax=2π3when9a2=49a2a=23b=3a=2

Answered by ajfour last updated on 07/Feb/21

let  x=pt  ⇒  y=3pt+2  let  3p=1  ⇒  (t^2 /(9a^2 ))+(y^2 /b^2 )=1  Now max area  when  r=3a=b  (√2)=r=b=3a  A_(max) =((2π)/3)

letx=pty=3pt+2let3p=1t29a2+y2b2=1Nowmaxareawhenr=3a=b2=r=b=3aAmax=2π3

Commented by ajfour last updated on 07/Feb/21

Commented by ajfour last updated on 07/Feb/21

3x=t

3x=t

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