What is the foci y=-7(x-8)^2+6 ?

The foci y=-7(x-8)^2+6 is (8, 167/28)

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foci y=-7(x-8)^2+6

foci

y = - 7 (x - 8)^{2} + 6

(8, \frac{167}{28})

Solution steps

Rewrite

y = - 7 (x - 8)^{2} + 6

in the standard form:

4 (- \frac{1}{28}) (y - 6) = (x - 8)^{2}

(h, k) = (8, 6), p = - \frac{1}{28}

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

p

from the center

(8, 6)

along the y-axis

(8, 6 + p)

= (8, 6 + (- \frac{1}{28}))

Refine

(8, \frac{167}{28})

96 t - 16 t^{2}

y = - 2 x^{2} + x - 1

4 x^{2} + 2 x + 4

y^{2} + 16 x = 0

y = 16 - x^{2}