What is the foci (y+1)^2=2(x-5) ?

The foci (y+1)^2=2(x-5) is (11/2 ,-1)

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foci (y+1)^2=2(x-5)

foci

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

(\frac{11}{2}, - 1)

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{2} (x - 5) = (y - (- 1))^{2}

(h, k) = (5, - 1), p = \frac{1}{2}

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

p

from the center

(5, - 1)

along the x-axis

(5 + p, - 1)

= (5 + \frac{1}{2}, - 1)

Refine

(\frac{11}{2}, - 1)

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

4 x = y^{2} - 4 y

3 x^{2} - 12 x + 2 y + 15 = 0

y^{2} - 3 y

y^{2} - y - 42