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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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- Thread starter amaresh92
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- #1

- 163

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what is the radius of curvature of a parabola y^2=4ax at the end of the focal chord ?

- #2

tiny-tim

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

(and

- #3

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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

so i'm

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