What is 16x^2+64x+4y^2-8y+4=0 ?

The solution to 16x^2+64x+4y^2-8y+4=0 is Ellipse with (h,k)=(-2,1),a=2,b=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16x^2+64x+4y^2-8y+4=0

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

(h, k) = (- 2, 1), a = 2, b = 4

Solution steps

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

Rewrite

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 1), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), b = 4, a = 2

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

49 x^{2} + 4 y^{2} = 196

18 y^{2} + 24 y + 12 x^{2} - 12 x = 1

x^{2} + 2 y^{2} - 2 x - 4 y = - 1

2 x^{2} + y^{2} \leq 1

What is 16x^2+64x+4y^2-8y+4=0 ?
The solution to 16x^2+64x+4y^2-8y+4=0 is Ellipse with (h,k)=(-2,1),a=2,b=4