What is the foci 2pir^2+16pir ?

The foci 2pir^2+16pir is (-4,(-256pi^2+1)/(8pi))

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foci 2pir^2+16pir

foci

2 π r^{2} + 16 π r

(- 4, \frac{- 256 π ^{2} + 1}{8 π})

Solution steps

Rewrite

y = 2 π r^{2} + 16 π r

in the standard form

4 \cdot \frac{1}{8 π} (y - (- 32 π)) = (r - (- 4))^{2}

(h, k) = (- 4, - 32 π), p = \frac{1}{8 π}

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

p

from the center

(- 4, - 32 π)

along the y-axis

(- 4, - 32 π + p)

= (- 4, - 32 π + \frac{1}{8 π})

Refine

(- 4, \frac{- 256 π ^{2} + 1}{8 π})

x^{2} - 4 x - 8 y - 12 = 0

(x - 6)^{2} = - 4 y

4 t^{2} + 1

t^{2} + 2 t - 48

x = - \frac{1}{2} (y - 4)^{2} + \frac{1}{2}

What is the foci 2pir^2+16pir ?
The foci 2pir^2+16pir is (-4,(-256pi^2+1)/(8pi))