What is 4y^{(2)}+2x^{(2)}+48y+4x-22=0 ?

The solution to 4y^{(2)}+2x^{(2)}+48y+4x-22=0 is Ellipse with (h,k)=(-1,-6),a=2sqrt(21),b=sqrt(42)

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4y^{(2)}+2x^{(2)}+48y+4x-22=0

4 y^{(2)} + 2 x^{(2)} + 48 y + 4 x - 22 = 0

(h, k) = (- 1, - 6), a = 221, b = 42

Solution steps

4 y^{2} + 2 x^{2} + 48 y + 4 x - 22 = 0

Rewrite

4 y^{2} + 2 x^{2} + 48 y + 4 x - 22 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( 2 21 ) ^{2}} + \frac{( y - ( - 6 ) ) ^{2}}{( 42 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, - 6), a = 221, b = 42

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 6), a = 221, b = 42

\frac{( x - 2 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

4 x^{2} + y^{2} - 24 x - 8 y + 8 = 0

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

4 x^{2} + y^{2} - 24 x + 2 y + 21 = 0

What is 4y^{(2)}+2x^{(2)}+48y+4x-22=0 ?
The solution to 4y^{(2)}+2x^{(2)}+48y+4x-22=0 is Ellipse with (h,k)=(-1,-6),a=2sqrt(21),b=sqrt(42)