What is the foci 10x^2-19x+5 ?

The foci 10x^2-19x+5 is (19/20 ,-4)

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foci 10x^2-19x+5

foci

10 x^{2} - 19 x + 5

(\frac{19}{20}, - 4)

Solution steps

Rewrite

y = 10 x^{2} - 19 x + 5

in the standard form

4 \cdot \frac{1}{40} (y - (- \frac{161}{40})) = (x - \frac{19}{20})^{2}

(h, k) = (\frac{19}{20}, - \frac{161}{40}), p = \frac{1}{40}

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

p

from the center

(\frac{19}{20}, - \frac{161}{40})

along the y-axis

(\frac{19}{20}, - \frac{161}{40} + p)

= (\frac{19}{20}, - \frac{161}{40} + \frac{1}{40})

Refine

(\frac{19}{20}, - 4)

+ (\frac{1}{2} x - 1)^{2} - 2

x^{2} + x - 5

x^{2} + x - 4

(x + 6)^{2} = 9 (y - 1)

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