What is the foci y=4x^2+12x+3 ?

The foci y=4x^2+12x+3 is (-3/2 ,-95/16)

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foci y=4x^2+12x+3

foci

y = 4 x^{2} + 12 x + 3

(- \frac{3}{2}, - \frac{95}{16})

Solution steps

Rewrite

y = 4 x^{2} + 12 x + 3

in the standard form

4 \cdot \frac{1}{16} (y - (- 6)) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, - 6), p = \frac{1}{16}

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

p

from the center

(- \frac{3}{2}, - 6)

along the y-axis

(- \frac{3}{2}, - 6 + p)

= (- \frac{3}{2}, - 6 + \frac{1}{16})

Refine

(- \frac{3}{2}, - \frac{95}{16})

y^{2} = x

y = 3 x^{2} - 12 x + 8

x = (y + 3)^{2}

x^{2} - 6 x + 5 y = 34

y = 2 x (x - 9) - (x - 9)