At what point of the parabola x²=9y is the abscissa three times that of ordinate ?

pls answer it as soon as possible.

The given parabola is x² = 9y.

Let P(h, k) be the point on the parabola such that abscissa is three times the ordinate.

∴ h = 3k    ...(1)

P(h, k) lies on the parabola x² = 9y,

∴ h² = 9k   ...(2)

From (1) and (2), we have

(3k)² = 9k

∴ 9k² – 9k = 0

⇒ 9k (k – 1) = 0

⇒ k = 0 or k – 1 = 0

⇒ k = 0 or k = 1

∴ k = 1                (When k = 0, h = 0)

When k = 1, we get

h = 3 × 1 = 3           (Using (1))

Thus, the required point is (3, 1).