What is the foci 5x^2-40x+4y+85=0 ?

The foci 5x^2-40x+4y+85=0 is (4,-29/20)

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foci 5x^2-40x+4y+85=0

foci

5 x^{2} - 40 x + 4 y + 85 = 0

(4, - \frac{29}{20})

Solution steps

Rewrite

5 x^{2} - 40 x + 4 y + 85 = 0

in the standard form

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

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

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

p

from the center

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

along the y-axis

(4, - \frac{5}{4} + p)

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

Refine

(4, - \frac{29}{20})

x^{2} - 3 x + 10

y = + x^{2} - x

x^{2} + 8 x - 16

y = + 3 x^{2} + 24 x - 51

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