What is 2x^2+2x^2-8x-12y+8=0 ?

The solution to 2x^2+2x^2-8x-12y+8=0 is Parabola with (h,k)=(1, 1/3),p= 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+2x^2-8x-12y+8=0

2 x^{2} + 2 x^{2} - 8 x - 12 y + 8 = 0

(h, k) = (1, \frac{1}{3}), p = \frac{3}{4}

Solution steps

2 x^{2} + 2 x^{2} - 8 x - 12 y + 8 = 0

Rewrite

2 x^{2} + 2 x^{2} - 8 x - 12 y + 8 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (1, \frac{1}{3}), p = \frac{3}{4}

- y = x^{2} + 2

- y = x^{2} + 1

x^{2} + 4 x + 8 y - 28 = 0

y \geq 3 x^{2} - 9 x + 6

y = - 2 x^{2} + 20 x - 52

What is 2x^2+2x^2-8x-12y+8=0 ?
The solution to 2x^2+2x^2-8x-12y+8=0 is Parabola with (h,k)=(1, 1/3),p= 3/4