What is 5(x+1)^2+3y^2=15 ?

The solution to 5(x+1)^2+3y^2=15 is Ellipse with (h,k)=(-1,0),a=sqrt(3),b=sqrt(5)

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5(x+1)^2+3y^2=15

5 (x + 1)^{2} + 3 y^{2} = 15

(h, k) = (- 1, 0), a = 3, b = 5

Solution steps

5 (x + 1)^{2} + 3 y^{2} = 15

Rewrite

5 (x + 1)^{2} + 3 y^{2} = 15

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( 3 ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 5 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), b = 5, a = 3

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

\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}

What is 5(x+1)^2+3y^2=15 ?
The solution to 5(x+1)^2+3y^2=15 is Ellipse with (h,k)=(-1,0),a=sqrt(3),b=sqrt(5)