What is the foci y=5x^2+6x-7 ?

The foci y=5x^2+6x-7 is (-3/5 ,-35/4)

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foci y=5x^2+6x-7

foci

y = 5 x^{2} + 6 x - 7

(- \frac{3}{5}, - \frac{35}{4})

Solution steps

Rewrite

y = 5 x^{2} + 6 x - 7

in the standard form

4 \cdot \frac{1}{20} (y - (- \frac{44}{5})) = (x - (- \frac{3}{5}))^{2}

(h, k) = (- \frac{3}{5}, - \frac{44}{5}), p = \frac{1}{20}

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

p

from the center

(- \frac{3}{5}, - \frac{44}{5})

along the y-axis

(- \frac{3}{5}, - \frac{44}{5} + p)

= (- \frac{3}{5}, - \frac{44}{5} + \frac{1}{20})

Refine

(- \frac{3}{5}, - \frac{35}{4})

2 x^{2} - 12 x + 3

9 y^{2} - 6 y - 9 - x = 0

x - x^{2}

x^{2} + 5 x + 6

x^{2} + 6 x + 16 y + 121 = 0