What is the foci x=5(y+3)^2+6 ?

The foci x=5(y+3)^2+6 is (121/20 ,-3)

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foci x=5(y+3)^2+6

foci

x = 5 (y + 3)^{2} + 6

(\frac{121}{20}, - 3)

Solution steps

Rewrite

x = 5 (y + 3)^{2} + 6

in the standard form:

4 \cdot \frac{1}{20} (x - 6) = (y - (- 3))^{2}

(h, k) = (6, - 3), p = \frac{1}{20}

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

p

from the center

(6, - 3)

along the x-axis

(6 + p, - 3)

= (6 + \frac{1}{20}, - 3)

Refine

(\frac{121}{20}, - 3)

y = \frac{3 \cdot \frac{1}{2} x ^{2}}{2}

5 x^{2} + 3

(x - 2)^{2} = 4 (y - 2)

y = 3 x^{2} - 6 x + 5

2 x^{2} - 12 x - 5