What is the foci x=-1/2 (y-2)^2+5 ?

The foci x=-1/2 (y-2)^2+5 is (9/2 ,2)

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foci x=-1/2 (y-2)^2+5

foci

x = - \frac{1}{2} (y - 2)^{2} + 5

(\frac{9}{2}, 2)

Solution steps

Rewrite

x = - \frac{1}{2} (y - 2)^{2} + 5

in the standard form:

4 (- \frac{1}{2}) (x - 5) = (y - 2)^{2}

(h, k) = (5, 2), p = - \frac{1}{2}

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

p

from the center

(5, 2)

along the x-axis

(5 + p, 2)

= (5 + (- \frac{1}{2}), 2)

Refine

(\frac{9}{2}, 2)

(x + 6)^{2} = - 8 (y + 9)

(2 + x)^{2}

\frac{40 x ^{3}}{3 x}

y = 10 x - x^{2}

y = (6 x + 7)^{2}