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Popular >

foci y^2=16x

Solution

foci

y^{2} = 16 x

Solution

(4, 0)

Solution steps

Rewrite

y^{2} = 16 x

in the standard form:

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

(h, k) = (0, 0), p = 4

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 + 4, 0)

Refine

(4, 0)

Popular Examples

(x-3)^2+(y+5)^2=16

(x - 3)^{2} + (y + 5)^{2} = 16

y^{2} - 8 x - 16 = 0

foci (x^2)/(16)-(y^2)/(25)=1 foci

\frac{x ^{2}}{16} - \frac{y ^{2}}{25} = 1

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

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

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

Frequently Asked Questions (FAQ)

What is the foci y^2=16x ?
The foci y^2=16x is (4,0)