What is the foci 2x^2+4x-y=0 ?

The foci 2x^2+4x-y=0 is (-1,-15/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

foci

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

(- 1, - \frac{15}{8})

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{8} (y - (- 2)) = (x - (- 1))^{2}

(h, k) = (- 1, - 2), p = \frac{1}{8}

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

p

from the center

(- 1, - 2)

along the y-axis

(- 1, - 2 + p)

= (- 1, - 2 + \frac{1}{8})

Refine

(- 1, - \frac{15}{8})

y^{2} - 12 x - 2 y - 25 = 0

y = 4 x^{2} + 8 x - 20

x = - \frac{1}{52} (y^{2} + 18 y - 647)

4 x^{2} + 8 x

x^{2} - 13 x + 40