What is x^2+4y^2+2x-16y+13=0 ?

The solution to x^2+4y^2+2x-16y+13=0 is Ellipse with (h,k)=(-1,2),a=2,b=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+4y^2+2x-16y+13=0

x^{2} + 4 y^{2} + 2 x - 16 y + 13 = 0

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

Solution steps

x^{2} + 4 y^{2} + 2 x - 16 y + 13 = 0

Rewrite

x^{2} + 4 y^{2} + 2 x - 16 y + 13 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

x^{2} + 4 y^{2} = 3

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{4} = 1

9 x^{2} + 4 y^{2} - 36 x - 8 y + 4 = 0

x^{2} + 3 y^{2} + 18 y + 18 = 0

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

What is x^2+4y^2+2x-16y+13=0 ?
The solution to x^2+4y^2+2x-16y+13=0 is Ellipse with (h,k)=(-1,2),a=2,b=1