What is 2x^2+3y^2+4x-1=0 ?

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

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2x^2+3y^2+4x-1=0

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

(h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

Solution steps

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

Rewrite

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( \frac{3}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

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

16 x^{2} + 16 y^{2} - 8 y^{2} - 15 = 0

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

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{36} = 1

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

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