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Popular >

foci y^2=20x

Solution

foci

y^{2} = 20 x

Solution

(5, 0)

Solution steps

Rewrite

y^{2} = 20 x

in the standard form:

4 \cdot 5 (x - 0) = (y - 0)^{2}

(h, k) = (0, 0), p = 5

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 + 5, 0)

Refine

(5, 0)

Popular Examples

(x^2)/(49)+(y^2)/(16)=1

\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1

((y-6)^2)/(64)-((x-8)^2)/(36)=1

\frac{( y - 6 ) ^{2}}{64} - \frac{( x - 8 ) ^{2}}{36} = 1

asymptotes of 9x^2-9y^2+18x+36y=63 asymptotes

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63

vertices f(x)=x^2+8x+12 vertices

f (x) = x^{2} + 8 x + 12

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

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

Frequently Asked Questions (FAQ)

What is the foci y^2=20x ?
The foci y^2=20x is (5,0)