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

The foci 2y=3x^2+x is (-1/6 , 1/8)

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foci 2y=3x^2+x

foci

2 y = 3 x^{2} + x

(- \frac{1}{6}, \frac{1}{8})

Solution steps

Rewrite

2 y = 3 x^{2} + x

in the standard form

4 \cdot \frac{1}{6} (y - (- \frac{1}{24})) = (x - (- \frac{1}{6}))^{2}

(h, k) = (- \frac{1}{6}, - \frac{1}{24}), p = \frac{1}{6}

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

p

from the center

(- \frac{1}{6}, - \frac{1}{24})

along the y-axis

(- \frac{1}{6}, - \frac{1}{24} + p)

= (- \frac{1}{6}, - \frac{1}{24} + \frac{1}{6})

Refine

(- \frac{1}{6}, \frac{1}{8})

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

y^{2} - 12 x - 2 y - 25 = 0

y = 4 x^{2} + 8 x - 20

x = - \frac{1}{52} (y^{2} + 18 y - 647)

4 x^{2} + 8 x