What is 4x^2+2y^2+6x+8y+4=0 ?

The solution to 4x^2+2y^2+6x+8y+4=0 is Ellipse with (h,k)=(-3/4 ,-2),a= 5/4 ,b= 5/(2sqrt(2))

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4x^2+2y^2+6x+8y+4=0

4 x^{2} + 2 y^{2} + 6 x + 8 y + 4 = 0

(h, k) = (- \frac{3}{4}, - 2), a = \frac{5}{4}, b = \frac{5}{2 2}

Solution steps

4 x^{2} + 2 y^{2} + 6 x + 8 y + 4 = 0

Rewrite

4 x^{2} + 2 y^{2} + 6 x + 8 y + 4 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{3}{4} ) ) ^{2}}{( \frac{5}{4} ) ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{( \frac{5}{2 2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{3}{4}, - 2), a = \frac{5}{4}, b = \frac{5}{2 2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{3}{4}, - 2), b = \frac{5}{2 2}, a = \frac{5}{4}

\frac{1 + b + ia}{1 + b - ia} = b + ia

2 x^{2} + y^{2} + 4 x - 6 y - 12 = 0

(\frac{X}{16})^{2} + \frac{( Y + 1 ) ^{2}}{4} - 1 = 0

\frac{( x - 0 ) ^{2}}{4} + \frac{( y - 0 ) ^{2}}{16} = 1

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

What is 4x^2+2y^2+6x+8y+4=0 ?
The solution to 4x^2+2y^2+6x+8y+4=0 is Ellipse with (h,k)=(-3/4 ,-2),a= 5/4 ,b= 5/(2sqrt(2))