What is the foci 2y^2-5y+x-4=0 ?

The foci 2y^2-5y+x-4=0 is (7, 5/4)

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foci 2y^2-5y+x-4=0

foci

2 y^{2} - 5 y + x - 4 = 0

(7, \frac{5}{4})

Solution steps

Rewrite

2 y^{2} - 5 y + x - 4 = 0

in the standard form

4 (- \frac{1}{8}) (x - \frac{57}{8}) = (y - \frac{5}{4})^{2}

(h, k) = (\frac{57}{8}, \frac{5}{4}), p = - \frac{1}{8}

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

p

from the center

(\frac{57}{8}, \frac{5}{4})

along the x-axis

(\frac{57}{8} + p, \frac{5}{4})

= (\frac{57}{8} + (- \frac{1}{8}), \frac{5}{4})

Refine

(7, \frac{5}{4})

2 y^{2} - 16 x = 0

y = - 3 x^{2}

y^{2} - 4 x + 6 y + 17 = 0

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

(x - 2)^{2} = 8 (y - 1)