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

The foci x^2+2x-y=3 is (-1,-15/4)

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foci x^2+2x-y=3

foci

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

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

Solution steps

Rewrite

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

in the standard form

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

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

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

p

from the center

(- 1, - 4)

along the y-axis

(- 1, - 4 + p)

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

Refine

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

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

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

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

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

4 x - x^{2}