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Popular >

vertices y^2=16x

Solution

vertices

y^{2} = 16 x

Solution

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

Solution steps

y^{2} = 16 x

Rewrite

y^{2} = 16 x

in the standard form:

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

Therefore parabola properties are:

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

Popular Examples

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

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

(x^2)/(a^2)+(y^2)/(b^2)=1

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

foci (x^2)/7-(y^2)/2 =1 foci

\frac{x ^{2}}{7} - \frac{y ^{2}}{2} = 1

x^2+y^2+12x-12y+36=0

x^{2} + y^{2} + 12 x - 12 y + 36 = 0

x^{2} + y^{2} = 18

Frequently Asked Questions (FAQ)

What is the vertices y^2=16x ?
The vertices y^2=16x is (h,k)=(0,0),p=4