What is 2x^2+3y^2+4x=19 ?

The solution to 2x^2+3y^2+4x=19 is Ellipse with (h,k)=(-1,0),a=sqrt(21/2),b=sqrt(7)

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2x^2+3y^2+4x=19

2 x^{2} + 3 y^{2} + 4 x = 19

(h, k) = (- 1, 0), a = \frac{21}{2}, b = 7

Solution steps

2 x^{2} + 3 y^{2} + 4 x = 19

Rewrite

2 x^{2} + 3 y^{2} + 4 x = 19

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( \frac{21}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 7 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = \frac{21}{2}, b = 7

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), a = \frac{21}{2}, b = 7

x^{2} + 6 x + (5 y + 6)^{2} - 2 (5 y + 6) + 10 = 4

25 y^{2} + 36 (x - 8)^{2} = 900

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

4 x^{2} + 3 y^{2} - 32 x - 18 y + 43 = 0

2 = \frac{y}{( y ^{2} + x ^{2} )}

What is 2x^2+3y^2+4x=19 ?
The solution to 2x^2+3y^2+4x=19 is Ellipse with (h,k)=(-1,0),a=sqrt(21/2),b=sqrt(7)