What is the foci x^2+6x-8y+1=0 ?

The foci x^2+6x-8y+1=0 is (-3,1)

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foci x^2+6x-8y+1=0

foci

x^{2} + 6 x - 8 y + 1 = 0

(- 3, 1)

Solution steps

Rewrite

x^{2} + 6 x - 8 y + 1 = 0

in the standard form

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

(h, k) = (- 3, - 1), p = 2

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

p

from the center

(- 3, - 1)

along the y-axis

(- 3, - 1 + p)

= (- 3, - 1 + 2)

Refine

(- 3, 1)

y = x^{2} + 1

290 (8 x + 0.5 x^{2} - 1.4)

y = - 3 x^{2} - 12 x - 9

2 - 3 x^{2}

x^{2} + 5 x - 14