Radius of curvature

  • Thread starter amaresh92
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  • #1
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what is the radius of curvature of a parabola y^2=4ax at the end of the focal chord ?
 

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  • #2
tiny-tim
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hi amaresh92! :wink:

show us what you've done, and where you're stuck, and then we'll know how to help! :smile:

(and which focal chord? do you mean the latus rectum? :confused:)
 
  • #3
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we got the formula,
r=[{(1+(dy/dx)^2)}^3/2]/(d2x/dy2)
we have,
y^2=4ax

2ydy/dx=4a
dy/dx=4a/2y=2/y

d2y/dx2=-2/y^2*dy/dx

d2y/dx2 =-2/y^2*2/y

d2y/dx2 =-4/y^3

r={(1+4/y^2)^3/2}/(-4/y^3)

then how to find the point at which it is asked?
its not a latus rectum i guess as it is mention only focal chord
 
  • #4
tiny-tim
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then how to find the point at which it is asked?
its not a latus rectum i guess as it is mention only focal chord

i've never heard of the term "focal chord" …

on that face of it, it should mean any chord through the focus …

so i'm guessing that they mean the latus rectum :confused:
 

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