What is the foci x=(2y-1)^2 ?

The foci x=(2y-1)^2 is (1/16 , 1/2)

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foci x=(2y-1)^2

foci

x = (2 y - 1)^{2}

(\frac{1}{16}, \frac{1}{2})

Solution steps

Rewrite

x = (2 y - 1)^{2}

in the standard form:

4 \cdot \frac{1}{16} (x - 0) = (y - \frac{1}{2})^{2}

(h, k) = (0, \frac{1}{2}), p = \frac{1}{16}

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

p

from the center

(0, \frac{1}{2})

along the x-axis

(0 + p, \frac{1}{2})

= (0 + \frac{1}{16}, \frac{1}{2})

Refine

(\frac{1}{16}, \frac{1}{2})

y = x^{2} - 10 x - 24

y = 92 x^{2}

y^{2} = 40 x

y = - 7 (x - 8)^{2} + 6

96 t - 16 t^{2}