A curve is such that the length of the polar radius of any pt. on the curve is equal to length of tangent at this pt.. Form diff. eqn. and find the curve ?

Dear Mayank

let a point on the curve is (x,y)

length of polar radius  r = √x²+y²

length of tangent  = [y√{1+(dy/dx)²} ]/(dy/dx) 

so   [y√{1+(dy/dx)²} ]/(dy/dx)  = √x²+y²

     square and simplyfy

  y=±xdy/dx

dx/x = ±dy/dx

intigrate

 lnx  =± lny  +lnC

+ve sign

x=Cy

-ve sign

x=C/y

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