Given v = (0,0) and axis is y-axis
The parabola is of the form x2 = 4ay or x2 = -4ay
But this equation passes through (-1,-0) which is in III qhadrant.
∴ The curve turn down words and is of the form x2 = – 4ay,
But it passes through (-1,-3)
∴ (-1)2 = -4a(-3) ⇒ 1 = 12a ⇒ a = \(\frac{1}{12}\)
∴ The required equation is x2 = -4 . \(\frac{1}{12}\) y
⇒ x2 = \(\frac{-1}{3}\) or 3x2 + y = 0