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The prependicular from the origin to the tangent at any pt. on curve is equal to abscissa of that pt Find curve ?
Dear Mayank
let the point on the curve is (x1,y1)
slope of the tangent at this point is (dy/dx)1
equation of tangent
y -y1 =(dy/dx)1 (x-x1)
y -x(dy/dx)1 +x1(dy/dx)1 -y1 =0
length of normal from origin
|x1(dy/dx)1 -y1| /√{(dy/dx)12 + 1} = |x1|
|x1(dy/dx)1 -y1| = |x1|√{(dy/dx)12 + 1}
square and simplyfy
{y12 -x12}/2x1y1 =(dydx)1
and for general point
{y2 -x2}/2xy =(dydx)
this is ordinary diferential equation you cab easly solve
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