What is the foci y=-x^2+4x+3 ?

The foci y=-x^2+4x+3 is (2, 27/4)

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foci y=-x^2+4x+3

foci

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

(2, \frac{27}{4})

Solution steps

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (y - 7) = (x - 2)^{2}

(h, k) = (2, 7), p = - \frac{1}{4}

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

p

from the center

(2, 7)

along the y-axis

(2, 7 + p)

= (2, 7 + (- \frac{1}{4}))

Refine

(2, \frac{27}{4})

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

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

x^{2} - 2 x - 4

x^{2} - 2 x - 3

x^{2} + 8 x + 18