What is the foci x=3(y-3)^2-1 ?

The foci x=3(y-3)^2-1 is (-11/12 ,3)

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foci x=3(y-3)^2-1

foci

x = 3 (y - 3)^{2} - 1

(- \frac{11}{12}, 3)

Solution steps

Rewrite

x = 3 (y - 3)^{2} - 1

in the standard form:

4 \cdot \frac{1}{12} (x - (- 1)) = (y - 3)^{2}

(h, k) = (- 1, 3), p = \frac{1}{12}

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

p

from the center

(- 1, 3)

along the x-axis

(- 1 + p, 3)

= (- 1 + \frac{1}{12}, 3)

Refine

(- \frac{11}{12}, 3)

(x + 2)^{2} - 3

y = 9 x^{2}

x = - \frac{1}{8} (y - 8)^{2} + 2

1 - x^{2}

25 x^{2} - 100