What is the foci y=4x^2-9 ?

The foci y=4x^2-9 is (0,-143/16)

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foci y=4x^2-9

foci

y = 4 x^{2} - 9

(0, - \frac{143}{16})

Solution steps

Rewrite

y = 4 x^{2} - 9

in the standard form:

4 \cdot \frac{1}{16} (y - (- 9)) = (x - 0)^{2}

(h, k) = (0, - 9), p = \frac{1}{16}

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

p

from the center

(0, - 9)

along the y-axis

(0, - 9 + p)

= (0, - 9 + \frac{1}{16})

Refine

(0, - \frac{143}{16})

y = 160 x - 16 x^{2}

(1060 - x)^{2} + x^{2}

y^{2} + 8 x = 0

9 x^{2} + 4 y = 1

x^{2} + 4 x + 4 y - 4 = 0