en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci y=8x^2-x+1

Solution

foci

y = 8 x^{2} - x + 1

Solution

(\frac{1}{16}, 1)

Solution steps

Rewrite

y = 8 x^{2} - x + 1

in the standard form

4 \cdot \frac{1}{32} (y - \frac{31}{32}) = (x - \frac{1}{16})^{2}

(h, k) = (\frac{1}{16}, \frac{31}{32}), p = \frac{1}{32}

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

p

from the center

(\frac{1}{16}, \frac{31}{32})

along the y-axis

(\frac{1}{16}, \frac{31}{32} + p)

= (\frac{1}{16}, \frac{31}{32} + \frac{1}{32})

Refine

(\frac{1}{16}, 1)

Popular Examples

foci 2y^2-5y+x-4=0 foci

2 y^{2} - 5 y + x - 4 = 0

foci 2y^2-16x=0 foci

2 y^{2} - 16 x = 0

foci y=-3x^2 foci

y = - 3 x^{2}

foci y^2-4x+6y+17=0 foci

y^{2} - 4 x + 6 y + 17 = 0

foci y=x^2-6x+5 foci

y = x^{2} - 6 x + 5

Frequently Asked Questions (FAQ)

What is the foci y=8x^2-x+1 ?
The foci y=8x^2-x+1 is (1/16 ,1)