What is the foci x^2+10x-2y+21=0 ?

The foci x^2+10x-2y+21=0 is (-5,-3/2)

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foci x^2+10x-2y+21=0

foci

x^{2} + 10 x - 2 y + 21 = 0

(- 5, - \frac{3}{2})

Solution steps

Rewrite

x^{2} + 10 x - 2 y + 21 = 0

in the standard form

4 \cdot \frac{1}{2} (y - (- 2)) = (x - (- 5))^{2}

(h, k) = (- 5, - 2), p = \frac{1}{2}

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

p

from the center

(- 5, - 2)

along the y-axis

(- 5, - 2 + p)

= (- 5, - 2 + \frac{1}{2})

Refine

(- 5, - \frac{3}{2})

y = 6 x^{2} + 5 x + 7

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

10 - x^{2}

x^{2} + 64

- 2 x^{2} - 12 x - 23