What is the foci y=(sqrt(x))^4+6(sqrt(x))^2-1 ?

The foci y=(sqrt(x))^4+6(sqrt(x))^2-1 is (-3,-39/4)

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foci y=(sqrt(x))^4+6(sqrt(x))^2-1

foci

y = (x)^{4} + 6 (x)^{2} - 1

(- 3, - \frac{39}{4})

Solution steps

Rewrite

y = (x)^{4} + 6 (x)^{2} - 1

in the standard form

4 \cdot \frac{1}{4} (y - (- 10)) = (x - (- 3))^{2}

(h, k) = (- 3, - 10), p = \frac{1}{4}

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

p

from the center

(- 3, - 10)

along the y-axis

(- 3, - 10 + p)

= (- 3, - 10 + \frac{1}{4})

Refine

(- 3, - \frac{39}{4})

y = 3 - 2 x - x^{2}

(y - 2)^{2} = 16 (x + 1)

6 x - 8 y + 15 - x^{2} = 0

4 x^{2} - 8 x + 1

x^{2} - 8 x + 2 y + 7 = 0

What is the foci y=(sqrt(x))^4+6(sqrt(x))^2-1 ?
The foci y=(sqrt(x))^4+6(sqrt(x))^2-1 is (-3,-39/4)