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

The foci-(x+3)^2-1 is (-3,-5/4)

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foci-(x+3)^2-1

foci

- (x + 3)^{2} - 1

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

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 3, - 1), p = - \frac{1}{4}

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 + (- \frac{1}{4}))

Refine

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

49 x^{2} + 16 y^{2} = 784

3 x^{2}

y^{2} - 8 x + 2 y + 17 = 0

x^{2} = - 36 y

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1