What is the foci y^2-6y-8x+17=0 ?

The foci y^2-6y-8x+17=0 is (3,3)

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foci y^2-6y-8x+17=0

foci

y^{2} - 6 y - 8 x + 17 = 0

(3, 3)

Solution steps

Rewrite

y^{2} - 6 y - 8 x + 17 = 0

in the standard form

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

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

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

p

from the center

(1, 3)

along the x-axis

(1 + p, 3)

= (1 + 2, 3)

Refine

(3, 3)

- 16 t^{2} + 256

2 x^{2} + 8 x - 24

10 y^{2} = 22 x

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

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