What is the foci y=2x^2+x-15 ?

The foci y=2x^2+x-15 is (-1/4 ,-15)

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foci y=2x^2+x-15

foci

y = 2 x^{2} + x - 15

(- \frac{1}{4}, - 15)

Solution steps

Rewrite

y = 2 x^{2} + x - 15

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{121}{8})) = (x - (- \frac{1}{4}))^{2}

(h, k) = (- \frac{1}{4}, - \frac{121}{8}), p = \frac{1}{8}

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

p

from the center

(- \frac{1}{4}, - \frac{121}{8})

along the y-axis

(- \frac{1}{4}, - \frac{121}{8} + p)

= (- \frac{1}{4}, - \frac{121}{8} + \frac{1}{8})

Refine

(- \frac{1}{4}, - 15)

x = y^{2} + 4 y + 10

(x + 7)^{2} = - 12 (y - 7)

3 y^{2} = - 4 x

x^{2} - 2 x - 8 y - 31 = 0

5 x^{2}