SOLUTION: Determine the equation of the parabola that has its vertex at the origin and satisfies the given conditions? Axis is the x-axis and p=2 Axis is the x-axis and the parabola pas

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Question 1055058: Determine the equation of the parabola that has its vertex at the origin and satisfies the given conditions?
Axis is the x-axis and p=2
Axis is the x-axis and the parabola passes through the point (4,2)
Write the equation in standard form.Determine the vertex ,axis and the direction in which each parabola opens
X^2-x+3y+1=0
X^2+2x+2y+3=0

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The first two example parts of the question take an equation form, 4p%28x-h%29=%28y-0%29%5E2, which you would be able to derive if you assumed a given vertex, directrix, focus. See a video about the derivation or the discussion in your textbook.



The example for p=2 would give 8%28x-h%29=%28y-0%29%5E2, and since you're also given that vertex is the origin, (0,0), the equation becomes simply 8x=y%5E2. If you want this in the more typical standard form, then x=highlight%28%281%2F8%29y%5E2%29.



The example for the parabola to contain point (4,2), and axis of symmetry still be x axis, means you have 4p%28x-h%29=%28y-0%29%5E2 or better, 4px=y%5E2; and you use the given included point to find the value of p.
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p=%28y%5E2%29%2F%284x%29
and putting in the coordinates for the point,
p=%282%5E2%29%2F%284%2A4%29
p=1%2F4
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and this finished equation is highlight%28x=y%5E2%29.