Sol. Let y = f(x) be the curve as shown in fig. and let P(x, y) be any point on the curve such that OL = OM. The equations of the tangent and normal at P are Y – y = (X –x) ... (i)

and Y – y = – (X –x) ... (ii) respectively.

It is given that the tangent and normal at P are equidistant from the origin.

∴ OL = OM ⇒ =

⇒ = ±

⇒ (y x) = (x ± y)

⇒ =

⇒ = or, =

Consider, =

It is a homogeneous differential equation.

Putting y = vx and = v + x we obtain

v + x =

⇒ x = –

⇒ dv = –

⇒ dv = –

⇒ dv + dv = – dx

⇒ log (v 2 + 1) + tan –1 v= –log x + log C

⇒ log = –tan –1 v

⇒ =

⇒ = C

Now, consider =

Putting y = vx and = v + x , we obtain

v + x =

⇒ x =

⇒ dv =

⇒ dv – . dv = dx

⇒ dv – dv = dx

⇒ tan –1 v – log (v 2 + 1) = log x + log C

⇒ tan –1 v = log {x(v 2 + 1) 1/2 C}

⇒ C(x 2 + y 2 ) 1/2 =

Hence, the equations of the curves are

= or, C =