(a) Given
focus is (0, -3) and directrix is y = 3
⇒ a = 3 and curve turns downwards and standard form is
x2 = -4ay put a = 3,
we get x2 = -12y
(b) Given focus S = (0,6) and Vertex, V = (0,0)
⇒ a = 6 and it is a upward parabola and the standard form is x2 = 4ay
Put a = 6
We get x2 = 24y
(c) Given vertex V = (0,0) and axis is y-axis and it passes through (\(\frac{1}{2}\), 2)
which is in I quadrant
∴ it is a upward parabola
x2 = 4ay
put x = \(\frac{1}{2}\) and y = 2; = 4a2 ⇒ \(\frac{1}{4}\) = 8a ⇒ a = \(\frac{1}{32}\)
∴ The required equation is x2 = 4ay
x2 =4. \(\frac{1}{32}\)y ⇒ x2 = \(\frac{1}{8}\)y or 8x2 – y = 0
(d) Vertex, V(0, 0) and y-axis and it passes through the point (-1, - 3)
which is in third quadrant
∴ It is a downward parabola,
i.e., x2 = -4ay
put x=-1 and y = -3; (-1)2 = -4a(-3)
1 = 12a ⇒ a = \(\frac{1}{12}\)
∴ Required equation is x2 = -4ay
x2 = -4. \(\frac{1}{12}\)y; 3x2 + y = 0.