What is the foci x^2+5y=14x+11 ?

The foci x^2+5y=14x+11 is (7, 43/4)

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foci x^2+5y=14x+11

foci

x^{2} + 5 y = 14 x + 11

(7, \frac{43}{4})

Solution steps

Rewrite

x^{2} + 5 y = 14 x + 11

in the standard form

4 (- \frac{5}{4}) (y - 12) = (x - 7)^{2}

(h, k) = (7, 12), p = - \frac{5}{4}

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

p

from the center

(7, 12)

along the y-axis

(7, 12 + p)

= (7, 12 + (- \frac{5}{4}))

Refine

(7, \frac{43}{4})

4 x^{2} + 16 x + 36

y = 4 x + 4 x - 2 + x^{2} - 3

y = (x - 5)^{2} - 9

25 x^{2} - 10 x - 200 y - 119 = 0

2 π r^{2} + 16 π r