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

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

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

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

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

Solution steps

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

Rewrite

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

78 = 32 (x - \frac{3}{2})^{2} + 50 (y + 1)^{2}

x^{2} + 2 y^{2} - 3 x + 8 y - 1 = 0

7 x^{2} = 35 - 5 y^{2}

4 x^{2} + 3 y^{2} = 25

4 x^{2} + 3 y^{2} = 26

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