What is the foci (y+1)^2=16(x-3) ?

The foci (y+1)^2=16(x-3) is (7,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (y+1)^2=16(x-3)

foci

(y + 1)^{2} = 16 (x - 3)

(7, - 1)

Solution steps

Rewrite

(y + 1)^{2} = 16 (x - 3)

in the standard form:

4 \cdot 4 (x - 3) = (y - (- 1))^{2}

(h, k) = (3, - 1), p = 4

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

p

from the center

(3, - 1)

along the x-axis

(3 + p, - 1)

= (3 + 4, - 1)

Refine

(7, - 1)

(x + 1) (x - 5)

(x - 6)^{2} = 1 (y - 6)

- x^{2} - 2

3 x^{2} + 8 y = 0

y^{2} - 6 x - y + 11 = 0