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in Parabola by (65.5k points)

Find the equation of the parabola given that its.

(a) Focus is (0, -3) and directrix is y = 3 

(b) Focus is (0, 6) and vertex is (0,0) 

(c) Vertex is (0,0), axis is y axis and passes through (\(\frac{1}{2}\), 2)

(d) Vertex is (0,0) axis is y axis and passes through (-1, -3)

1 Answer

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Best answer

(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.

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