What is x^2+4y^2+2x-12y+7=0 ?

The solution to x^2+4y^2+2x-12y+7=0 is Ellipse with (h,k)=(-1, 3/2),a=sqrt(3),b=(sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2+4y^2+2x-12y+7=0

x^{2} + 4 y^{2} + 2 x - 12 y + 7 = 0

(h, k) = (- 1, \frac{3}{2}), a = 3, b = \frac{3}{2}

Solution steps

x^{2} + 4 y^{2} + 2 x - 12 y + 7 = 0

Rewrite

x^{2} + 4 y^{2} + 2 x - 12 y + 7 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, \frac{3}{2}), a = 3, b = \frac{3}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, \frac{3}{2}), a = 3, b = \frac{3}{2}

What is x^2+4y^2+2x-12y+7=0 ?
The solution to x^2+4y^2+2x-12y+7=0 is Ellipse with (h,k)=(-1, 3/2),a=sqrt(3),b=(sqrt(3))/2