It is possible to rewrite this equation using the form $latex {{x}^2}=4py$, where p is a constant used to find the focus and directrix. This corresponds to a parabola with a vertical orientation.
As per this definition of the eccentricity of the parabola, we have PF = PB (Since e = PF/PB = 1) The coordinates of the focus is F (a,0) and we can use the coordinate distance formula to find its distance from P (x, y) PF = √(x−a)2
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How do I solve for p in parabolas? .. Standard forms for parabolas: x^2=4py and y^2=4px, with vertices at (0,0) or (x-h)^2=4p (y-k) and (y-k)^2=4p (x-h), with vertices at (h,k) The first
[math]\displaystyle y = ax^2 + bx + c \, [/math] To find p, use the following equation. [math]\displaystyle \frac {1} {4a} = p [/math] P allows you to find the focus and directrix, of you
STANDARD EQUATION OF A PARABOLA: Let the vertex be (h, k) and p be the distance between the vertex and the focus and p ≠ 0. (x − h) 2 = 4 p (y − k) vertical axis; directrix is y = k - p (y − k) 2 = 4 p (x − h) horizontal axis; directrix is x = h - p
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