What is the foci 2x^2-16x-3y+38=0 ?

The foci 2x^2-16x-3y+38=0 is (4, 19/8)

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foci 2x^2-16x-3y+38=0

foci

2 x^{2} - 16 x - 3 y + 38 = 0

(4, \frac{19}{8})

Solution steps

Rewrite

2 x^{2} - 16 x - 3 y + 38 = 0

in the standard form

4 \cdot \frac{3}{8} (y - 2) = (x - 4)^{2}

(h, k) = (4, 2), p = \frac{3}{8}

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

p

from the center

(4, 2)

along the y-axis

(4, 2 + p)

= (4, 2 + \frac{3}{8})

Refine

(4, \frac{19}{8})

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

x^{2} - 5 x - 7

- x^{2} + 4

y = 7 x^{2}

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