What is the foci-((x+3)^2)/(20)+1 ?

The foci-((x+3)^2)/(20)+1 is (-3,-4)

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foci-((x+3)^2)/(20)+1

foci

- \frac{( x + 3 ) ^{2}}{20} + 1

(- 3, - 4)

Solution steps

Rewrite

y = - \frac{( x + 3 ) ^{2}}{20} + 1

in the standard form:

4 (- 5) (y - 1) = (x - (- 3))^{2}

(h, k) = (- 3, 1), p = - 5

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

p

from the center

(- 3, 1)

along the y-axis

(- 3, 1 + p)

= (- 3, 1 + (- 5))

Refine

(- 3, - 4)

y = 3 x^{2} - 15 x + 6

- 12 y + 10 = x^{2} - 4 x + 14

(x + 4)^{2} = - 6 (y - 3)

(y + 1)^{2} = 2 (x - 5)

y = 5 x^{2} - 12 x + 7