What is the foci 9x^2+4y=1 ?

The foci 9x^2+4y=1 is (0, 5/36)

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foci 9x^2+4y=1

foci

9 x^{2} + 4 y = 1

(0, \frac{5}{36})

Solution steps

Rewrite

9 x^{2} + 4 y = 1

in the standard form:

4 (- \frac{1}{9}) (y - \frac{1}{4}) = (x - 0)^{2}

(h, k) = (0, \frac{1}{4}), p = - \frac{1}{9}

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

p

from the center

(0, \frac{1}{4})

along the y-axis

(0, \frac{1}{4} + p)

= (0, \frac{1}{4} + (- \frac{1}{9}))

Refine

(0, \frac{5}{36})

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

y = - \frac{1}{8} x^{2}

x^{2} + 16

(y - 3)^{2} = - 20 (x + 2)

- x^{2} + 6 x