What is x^2+4y^2+4x+32y-64=0 ?

The solution to x^2+4y^2+4x+32y-64=0 is Ellipse with (h,k)=(-2,-4),a=2sqrt(33),b=sqrt(33)

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x^2+4y^2+4x+32y-64=0

x^{2} + 4 y^{2} + 4 x + 32 y - 64 = 0

(h, k) = (- 2, - 4), a = 233, b = 33

Solution steps

x^{2} + 4 y^{2} + 4 x + 32 y - 64 = 0

Rewrite

x^{2} + 4 y^{2} + 4 x + 32 y - 64 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{( 2 33 ) ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{( 33 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, - 4), a = 233, b = 33

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, - 4), a = 233, b = 33

\frac{( x - 2 ) ^{2}}{10} + \frac{( y + 1 ) ^{2}}{25} = 1

36 x^{2} + 16 y^{2} + 144 x - 160 y - 32 = 0

9 x^{2} + 49 y^{2} - 114 x + 490 y + 1360 = 0

x^{2} + 4 z^{2} = 4

\frac{( x - 2 ) ^{2}}{25} + \frac{( v + 3 ) ^{2}}{9} = 1

What is x^2+4y^2+4x+32y-64=0 ?
The solution to x^2+4y^2+4x+32y-64=0 is Ellipse with (h,k)=(-2,-4),a=2sqrt(33),b=sqrt(33)