What is the foci x=-1/52 (y^2+18y-647) ?

The foci x=-1/52 (y^2+18y-647) is (1,-9)

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foci x=-1/52 (y^2+18y-647)

foci

x = - \frac{1}{52} (y^{2} + 18 y - 647)

(1, - 9)

Solution steps

Rewrite

x = - \frac{1}{52} (y^{2} + 18 y - 647)

in the standard form

4 (- 13) (x - 14) = (y - (- 9))^{2}

(h, k) = (14, - 9), p = - 13

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

p

from the center

(14, - 9)

along the x-axis

(14 + p, - 9)

= (14 + (- 13), - 9)

Refine

(1, - 9)

4 x^{2} + 8 x

x^{2} - 13 x + 40

- 2 x^{2} - 8 x - 12

\frac{4 x ^{3} - 7 x ^{2} - 11 x + 5}{4 x + 5}

y = 3 x^{2} - 5 x

What is the foci x=-1/52 (y^2+18y-647) ?
The foci x=-1/52 (y^2+18y-647) is (1,-9)