What is the foci y=(0.5-x)^2 ?

The foci y=(0.5-x)^2 is (0.5, 1/4)

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foci y=(0.5-x)^2

foci

y = (0.5 - x)^{2}

(0.5, \frac{1}{4})

Solution steps

Rewrite

y = (0.5 - x)^{2}

in the standard form:

4 \cdot \frac{1}{4} (y - 0) = (x - 0.5)^{2}

(h, k) = (0.5, 0), p = \frac{1}{4}

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

p

from the center

(0.5, 0)

along the y-axis

(0.5, 0 + p)

= (0.5, 0 + \frac{1}{4})

Refine

(0.5, \frac{1}{4})

25 x^{2} - 4

2 x^{2} - 8 y = 6 x^{2} + 20 y

y = 3 x^{2} - 35 x

y^{2} - 6 y - 2 x + 7 = 0

y = - 0.0000848 x^{2} - 0.006005 x + 1.004