What is the foci F=(mv^2)/r ?

The foci F=(mv^2)/r is (0, 1/(4m))

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foci F=(mv^2)/r

foci

F = \frac{m v ^{2}}{r}

(0, \frac{1}{4 m})

Solution steps

Rewrite

F - \frac{m v ^{2}}{r} = 0

in the standard form

4 \cdot \frac{1}{4 m} (r - 0) = (v - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{4 m}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(0, 0)

along the y-axis

(0, 0 + p)

= (0, 0 + \frac{1}{4 m})

Refine

(0, \frac{1}{4 m})

(x - 2)^{2} = 4 (y + 2)

(x - 2)^{2} = 4 (y + 3)

2 y^{2} - 6 x = 0

x^{2} - 2 x + 1

x^{2} + 9 x - 8