What is 4x^2-12x-16y+41=0 ?

The solution to 4x^2-12x-16y+41=0 is Parabola with (h,k)=(3/2 ,2),p=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2-12x-16y+41=0

4 x^{2} - 12 x - 16 y + 41 = 0

(h, k) = (\frac{3}{2}, 2), p = 1

Solution steps

4 x^{2} - 12 x - 16 y + 41 = 0

Rewrite

4 x^{2} - 12 x - 16 y + 41 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (\frac{3}{2}, 2), p = 1

2 y^{2} - 8 x - 24 y + 56 = 0

\frac{( x - 30 ) ^{2}}{30} - \frac{( y - 200 ) ^{2}}{50} = 3

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

x^{2} + y^{2} - 12 y + 32 = 0

What is 4x^2-12x-16y+41=0 ?
The solution to 4x^2-12x-16y+41=0 is Parabola with (h,k)=(3/2 ,2),p=1