What is the foci x=3-5(y+1)^2 ?

The foci x=3-5(y+1)^2 is (59/20 ,-1)

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foci x=3-5(y+1)^2

foci

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

(\frac{59}{20}, - 1)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(3, - 1)

along the x-axis

(3 + p, - 1)

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

Refine

(\frac{59}{20}, - 1)

y = x^{2} - 6 x + 11

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

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

- 16 x + y^{2} + 2 y - 63 = 0

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