We have y2 = x
⇒ dy/dx = 1/2y
Let the point P(t2, t) **** on the curve y2 = x
Equation of the normal to the curve at P is
y – t = –2t(x – t2)
which passes through the point (c, 0)
So, – t = – 2t(c – t2)
⇒ (c – t2) = 1/2
⇒ c – 1/2 = t 2
⇒ c – 1/2 > 0
⇒ c > 1/2
Hence, the value of c is c ∈u2009(1/2, ∞)