What is the foci (x+3)^2-2 ?

The foci (x+3)^2-2 is (-3,-7/4)

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foci (x+3)^2-2

foci

(x + 3)^{2} - 2

(- 3, - \frac{7}{4})

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{4} (y - (- 2)) = (x - (- 3))^{2}

(h, k) = (- 3, - 2), p = \frac{1}{4}

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

p

from the center

(- 3, - 2)

along the y-axis

(- 3, - 2 + p)

= (- 3, - 2 + \frac{1}{4})

Refine

(- 3, - \frac{7}{4})

5 x + 3 y^{2} = 0

y = 2 x^{2} - 3 x 8

12 + 4 x - x^{2}

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

5 x^{2} + 30 x + 24 y = 51