What is the foci x^2-4x-5y+7=0 ?

The foci x^2-4x-5y+7=0 is (2, 37/20)

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foci x^2-4x-5y+7=0

foci

x^{2} - 4 x - 5 y + 7 = 0

(2, \frac{37}{20})

Solution steps

Rewrite

x^{2} - 4 x - 5 y + 7 = 0

in the standard form

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

(h, k) = (2, \frac{3}{5}), p = \frac{5}{4}

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

p

from the center

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

along the y-axis

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

= (2, \frac{3}{5} + \frac{5}{4})

Refine

(2, \frac{37}{20})

(x - 2)^{2} = 12 (y - 1)

x^{2} - 2 x - 4

x^{2} - 2 x - 3

x^{2} + 8 x + 18

x^{2} + 8 x + 15