SOLUTION: A parabola has vertex (0,0) and focus on the y axis a) find the equation of the parabola if it passes through the point (4,12) Can you please help me out, thanks so much in adv

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A parabola has vertex (0,0) and focus on the y axis a) find the equation of the parabola if it passes through the point (4,12) Can you please help me out, thanks so much in adv      Log On


   



Question 800849: A parabola has vertex (0,0) and focus on the y axis
a) find the equation of the parabola if it passes through the point (4,12)
Can you please help me out, thanks so much in advance:)

Found 2 solutions by stanbon, DrBeeee:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A parabola has vertex (0,0) and focus on the y axis
a) find the equation of the parabola if it passes through the point (4,12)
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Plot those points and you will see the parabola opens up.
Form: (x-h)^2 = p(y-k)
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Using the 2 points you get:
(4-0)^2 = p(12-0)
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16 = 12p
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p = 4/3
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Equation: x^2 = (4/3)y
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y = (3/4)x^2
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Cheers,
Stan H.
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Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Since your parabola has its turning point at the origin, (0,0), and focus on y axis the equation is
(1)+y+=+a%2Ax%5E2+%2B0+%2B0+
I include the two zeroes in the equations to indicate that the y-intercept is zero and the axis of symmetry is at x equals zero.
Since the parabola passes through the point (4,12) we have
(2) 12 = a*4^2 or
(3) 12 = 16a or
(4) a = 12/16 or
(5) a = 3/4
Let's check this.
Is (12 = 3/4*4^2)?
Is (12 = 3/4*16)?
Is (12 = 3*4)?
Is (12 = 12)? Yes
Answer: The equation of the parabola that meets the conditions stated is y+=+%283%2F4%29x%5E2