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Question-208755




Question Number 208755 by efronzo1 last updated on 22/Jun/24
Answered by mr W last updated on 22/Jun/24
Commented by mr W last updated on 22/Jun/24
y=−(x^2 /2)+4x=−(((x−4)^2 )/2)+8  y=−(x^2 /2)+8 in new system (red)  touching point at (p, q)  q=−(p^2 /2)+8  −(1/(tan θ))=(dy/dx)=−p  ⇒tan θ=(1/p)  p=r cos θ=((rp)/( (√(1+p^2 )))) ⇒r=(√(1+p^2 ))  q=r+r sin θ=r(1+(1/( (√(1+p^2 )))))=−(p^2 /2)+8  (√(1+p^2 ))(1+(1/( (√(1+p^2 )))))=−(p^2 /2)+8  2(√(1+p^2 ))=−p^2 +14  p^4 −32p^2 +192=0  ⇒p^2 =16−8=8  r=(√(1+8))=3 ✓
y=x22+4x=(x4)22+8y=x22+8innewsystem(red)touchingpointat(p,q)q=p22+81tanθ=dydx=ptanθ=1pp=rcosθ=rp1+p2r=1+p2q=r+rsinθ=r(1+11+p2)=p22+81+p2(1+11+p2)=p22+821+p2=p2+14p432p2+192=0p2=168=8r=1+8=3
Commented by mr W last updated on 22/Jun/24
Commented by efronzo1 last updated on 23/Jun/24

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