What is the foci y=160x-16x^2 ?

The foci y=160x-16x^2 is (5, 25599/64)

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foci y=160x-16x^2

foci

y = 160 x - 16 x^{2}

(5, \frac{25599}{64})

Solution steps

Rewrite

y = 160 x - 16 x^{2}

in the standard form

4 (- \frac{1}{64}) (y - 400) = (x - 5)^{2}

(h, k) = (5, 400), p = - \frac{1}{64}

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

p

from the center

(5, 400)

along the y-axis

(5, 400 + p)

= (5, 400 + (- \frac{1}{64}))

Refine

(5, \frac{25599}{64})

(1060 - x)^{2} + x^{2}

y^{2} + 8 x = 0

9 x^{2} + 4 y = 1

x^{2} + 4 x + 4 y - 4 = 0

y = - \frac{1}{8} x^{2}