What is the foci y=6x^2-29x-5 ?

The foci y=6x^2-29x-5 is (29/12 ,-40)

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foci y=6x^2-29x-5

foci

y = 6 x^{2} - 29 x - 5

(\frac{29}{12}, - 40)

Solution steps

Rewrite

y = 6 x^{2} - 29 x - 5

in the standard form

4 \cdot \frac{1}{24} (y - (- \frac{961}{24})) = (x - \frac{29}{12})^{2}

(h, k) = (\frac{29}{12}, - \frac{961}{24}), p = \frac{1}{24}

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

p

from the center

(\frac{29}{12}, - \frac{961}{24})

along the y-axis

(\frac{29}{12}, - \frac{961}{24} + p)

= (\frac{29}{12}, - \frac{961}{24} + \frac{1}{24})

Refine

(\frac{29}{12}, - 40)

y = 1 - 3 x - x^{2}

x^{2} - 2 x - 4 y + 13 = 0

2 y^{2} - 16 y + 72 = 20 x

y^{2} + 4 x - 8 y + 12 = 0

- x^{2} + 4 x - 1