What is the foci x=y^2-4y+1 ?

The foci x=y^2-4y+1 is (-11/4 ,2)

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foci x=y^2-4y+1

foci

x = y^{2} - 4 y + 1

(- \frac{11}{4}, 2)

Solution steps

Rewrite

x = y^{2} - 4 y + 1

in the standard form

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

(h, k) = (- 3, 2), p = \frac{1}{4}

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

p

from the center

(- 3, 2)

along the x-axis

(- 3 + p, 2)

= (- 3 + \frac{1}{4}, 2)

Refine

(- \frac{11}{4}, 2)

\frac{- 3 x ^{2} - 2 x + 6}{5}

(y - 3)^{2} = - 12 x

(x - 2) = 3 (y - 5)^{2}

(x - 3)^{2} = 8 (y - 5)

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