What is the foci y^2= 16/3 x ?

The foci y^2= 16/3 x is (4/3 ,0)

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foci y^2= 16/3 x

foci

y^{2} = \frac{16}{3} x

(\frac{4}{3}, 0)

Solution steps

Rewrite

y^{2} = \frac{16}{3} x

in the standard form:

4 \cdot \frac{4}{3} (x - 0) = (y - 0)^{2}

(h, k) = (0, 0), p = \frac{4}{3}

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{4}{3}, 0)

Refine

(\frac{4}{3}, 0)

\frac{( x - 3 ) ^{2}}{8} + 5

(x + 5)^{2} + 3

- 0.1 x^{2} + 1.2 x + 98.6

y^{2} - 8 x - 6 y - 3 = 0

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