What is the foci (x+3)^2=12(y-5) ?

The foci (x+3)^2=12(y-5) is (-3,8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (x+3)^2=12(y-5)

foci

(x + 3)^{2} = 12 (y - 5)

(- 3, 8)

Solution steps

Rewrite

(x + 3)^{2} = 12 (y - 5)

in the standard form:

4 \cdot 3 (y - 5) = (x - (- 3))^{2}

(h, k) = (- 3, 5), p = 3

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

p

from the center

(- 3, 5)

along the y-axis

(- 3, 5 + p)

= (- 3, 5 + 3)

Refine

(- 3, 8)

x^{2} - 6 x

x^{2} - 3 x

x^{2} - 10 x + 15

2 x^{2} + 3

y = x^{2} + 9 x + 20