What is the foci 4x^2-4x+21 ?

The foci 4x^2-4x+21 is (1/2 , 321/16)

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foci 4x^2-4x+21

foci

4 x^{2} - 4 x + 21

(\frac{1}{2}, \frac{321}{16})

Solution steps

Rewrite

y = 4 x^{2} - 4 x + 21

in the standard form

4 \cdot \frac{1}{16} (y - 20) = (x - \frac{1}{2})^{2}

(h, k) = (\frac{1}{2}, 20), p = \frac{1}{16}

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

p

from the center

(\frac{1}{2}, 20)

along the y-axis

(\frac{1}{2}, 20 + p)

= (\frac{1}{2}, 20 + \frac{1}{16})

Refine

(\frac{1}{2}, \frac{321}{16})

y = (x + 314)^{2}

(x - 1)^{2} + 2

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

x^{2} + 14 x - 8 y + 65 = 0

y = 3 x^{2}