What is the foci x=y^2+4y+10 ?

The foci x=y^2+4y+10 is (25/4 ,-2)

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foci x=y^2+4y+10

foci

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

(\frac{25}{4}, - 2)

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (6, - 2), p = \frac{1}{4}

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

p

from the center

(6, - 2)

along the x-axis

(6 + p, - 2)

= (6 + \frac{1}{4}, - 2)

Refine

(\frac{25}{4}, - 2)

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

3 y^{2} = - 4 x

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

5 x^{2}

y = x^{2} - x + 1