What is the foci x=-3y^2+5 ?

The foci x=-3y^2+5 is (59/12 ,0)

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foci x=-3y^2+5

foci

x = - 3 y^{2} + 5

(\frac{59}{12}, 0)

Solution steps

Rewrite

x = - 3 y^{2} + 5

in the standard form:

4 (- \frac{1}{12}) (x - 5) = (y - 0)^{2}

(h, k) = (5, 0), p = - \frac{1}{12}

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

p

from the center

(5, 0)

along the x-axis

(5 + p, 0)

= (5 + (- \frac{1}{12}), 0)

Refine

(\frac{59}{12}, 0)

x^{2} + 4 x + 4 y = - 4

6 x^{2} - 24 x - 30

x^{2} + y = 4

y = x^{2} - 4 x - 12

x^{2} = - y