What is the foci y=9x^2+24x+16 ?

The foci y=9x^2+24x+16 is (-4/3 , 1/36)

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foci y=9x^2+24x+16

foci

y = 9 x^{2} + 24 x + 16

(- \frac{4}{3}, \frac{1}{36})

Solution steps

Rewrite

y = 9 x^{2} + 24 x + 16

in the standard form

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

(h, k) = (- \frac{4}{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{4}{3}, 0)

along the y-axis

(- \frac{4}{3}, 0 + p)

= (- \frac{4}{3}, 0 + \frac{1}{36})

Refine

(- \frac{4}{3}, \frac{1}{36})

x^{2} - 5 y = 0

x^{2} + 6 x - 1

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

y = x^{2} + 8 x - 5

y^{2} + 4 x - 14 y = - 53