What is the foci x^2-6x+5y=34 ?

The foci x^2-6x+5y=34 is (3, 147/20)

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foci x^2-6x+5y=34

foci

x^{2} - 6 x + 5 y = 34

(3, \frac{147}{20})

Solution steps

Rewrite

x^{2} - 6 x + 5 y = 34

in the standard form

4 (- \frac{5}{4}) (y - \frac{43}{5}) = (x - 3)^{2}

(h, k) = (3, \frac{43}{5}), p = - \frac{5}{4}

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

p

from the center

(3, \frac{43}{5})

along the y-axis

(3, \frac{43}{5} + p)

= (3, \frac{43}{5} + (- \frac{5}{4}))

Refine

(3, \frac{147}{20})

y = 2 x (x - 9) - (x - 9)

- 0.0000848 x^{2} - 0.006005 x + 1.004

x^{2} + 16 y = 0

y = 4 x^{2} - 9

y = 160 x - 16 x^{2}