What is the foci 10y^2=22x ?

The foci 10y^2=22x is (11/20 ,0)

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foci 10y^2=22x

foci

10 y^{2} = 22 x

(\frac{11}{20}, 0)

Solution steps

Rewrite

10 y^{2} = 22 x

in the standard form:

4 \cdot \frac{11}{20} (x - 0) = (y - 0)^{2}

(h, k) = (0, 0), p = \frac{11}{20}

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

p

from the center

(0, 0)

along the x-axis

(0 + p, 0)

= (0 + \frac{11}{20}, 0)

Refine

(\frac{11}{20}, 0)

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

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

x^{2} = - 10 y

y = 3 x^{2} + 5 x - 4

x^{2} - 4 x + 4