What is the foci y=(2x-1)(4x-1) ?

The foci y=(2x-1)(4x-1) is (3/8 ,-3/32)

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foci y=(2x-1)(4x-1)

foci

y = (2 x - 1) (4 x - 1)

(\frac{3}{8}, - \frac{3}{32})

Solution steps

Rewrite

y = (2 x - 1) (4 x - 1)

in the standard form

4 \cdot \frac{1}{32} (y - (- \frac{1}{8})) = (x - \frac{3}{8})^{2}

(h, k) = (\frac{3}{8}, - \frac{1}{8}), p = \frac{1}{32}

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

p

from the center

(\frac{3}{8}, - \frac{1}{8})

along the y-axis

(\frac{3}{8}, - \frac{1}{8} + p)

= (\frac{3}{8}, - \frac{1}{8} + \frac{1}{32})

Refine

(\frac{3}{8}, - \frac{3}{32})

(y + 6)^{2} = 12 (x + 1)

(x + 7)^{2} = 5 (y + 3)

x^{2} - 8 y = 0

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

- \frac{x ^{2}}{8} + \frac{x}{4} + \frac{23}{8}