What is the foci 4x^2+48y+12x=159 ?

The foci 4x^2+48y+12x=159 is (-3/2 , 1/2)

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foci 4x^2+48y+12x=159

foci

4 x^{2} + 48 y + 12 x = 159

(- \frac{3}{2}, \frac{1}{2})

Solution steps

Rewrite

4 x^{2} + 48 y + 12 x = 159

in the standard form

4 (- 3) (y - \frac{7}{2}) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, \frac{7}{2}), p = - 3

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

p

from the center

(- \frac{3}{2}, \frac{7}{2})

along the y-axis

(- \frac{3}{2}, \frac{7}{2} + p)

= (- \frac{3}{2}, \frac{7}{2} + (- 3))

Refine

(- \frac{3}{2}, \frac{1}{2})

y = 2 x^{2} + 5 x - 3

(x + 9)^{2} = - 40 (y - 7)

y = x^{2} - 14 x - 49

4 x - x^{2}

\frac{1}{2} (1 - 2 y - y^{2})