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

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

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

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

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

Solution steps

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

Rewrite

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

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 x^{2} + 8 y^{2} + 12 x + 48 y + 60 = 0

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

9 x^{2} + 4 y^{2} - 36 x + 24 y - 1 = 0

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

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

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