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

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

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

foci

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

(\frac{1}{2}, - 3)

Solution steps

Rewrite

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

in the standard form:

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

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

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}{2}), - 3)

Refine

(\frac{1}{2}, - 3)

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

4 x^{2} + 9 y^{2} = 36

3 x^{2} + 6 y^{2} = 18

x^{2} = - 12 y

y^{2} = 4 x