What is the foci y+4x<= x^2+2 ?

The foci y+4x<= x^2+2 is (2,-7/4)

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foci y+4x<= x^2+2

foci

y + 4 x \leq x^{2} + 2

(2, - \frac{7}{4})

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (2, - 2), p = \frac{1}{4}

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

p

from the center

(2, - 2)

along the y-axis

(2, - 2 + p)

= (2, - 2 + \frac{1}{4})

Refine

(2, - \frac{7}{4})

x^{2} - 6 x - 4 y + 5 = 0

y = 6 x^{2} + 26 x + 8

y = x^{2} + 7 x + 10

y = (2 x - 1) (4 x - 1)

(y + 6)^{2} = 12 (x + 1)