What is the foci y=9x^2+48x+64 ?

The foci y=9x^2+48x+64 is (-8/3 , 1/36)

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foci y=9x^2+48x+64

foci

y = 9 x^{2} + 48 x + 64

(- \frac{8}{3}, \frac{1}{36})

Solution steps

Rewrite

y = 9 x^{2} + 48 x + 64

in the standard form

4 \cdot \frac{1}{36} (y - 0) = (x - (- \frac{8}{3}))^{2}

(h, k) = (- \frac{8}{3}, 0), p = \frac{1}{36}

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

p

from the center

(- \frac{8}{3}, 0)

along the y-axis

(- \frac{8}{3}, 0 + p)

= (- \frac{8}{3}, 0 + \frac{1}{36})

Refine

(- \frac{8}{3}, \frac{1}{36})

x^{2} - 12 x + 35

4.905 x^{2} + 8 x - 30

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

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

(2 + x)^{2}