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

The foci x=(2y+5)^2 is (1/16 ,-5/2)

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foci x=(2y+5)^2

foci

x = (2 y + 5)^{2}

(\frac{1}{16}, - \frac{5}{2})

Solution steps

Rewrite

x = (2 y + 5)^{2}

in the standard form:

4 \cdot \frac{1}{16} (x - 0) = (y - (- \frac{5}{2}))^{2}

(h, k) = (0, - \frac{5}{2}), p = \frac{1}{16}

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

p

from the center

(0, - \frac{5}{2})

along the x-axis

(0 + p, - \frac{5}{2})

= (0 + \frac{1}{16}, - \frac{5}{2})

Refine

(\frac{1}{16}, - \frac{5}{2})

y = 8 x^{2}

y = 6 x - x^{2} - 5

(y + 3)^{2} = 4 (x - 3)

y = - 2 x^{2} + 24 x

(x - 1)^{2} = 8 (y + 3)