Identifying hyperbola vertices and foci from y^2+2y=4x^2+3

[StudentofSci]

Homework Statement

Identify the type of conic section whose equation is given as y^2+2y=4x^2+3
Also find its vertices and foci
(y-k)^2/a^2-(x-h)^2/b^2=1

Homework Equations

I believe when you have x&y ^2 in an equation it can be either an ellipse or a hyperbola but if the signs are opposite it is a hyperbola. vertices ± a from center, foci ± c from center.

The Attempt at a Solution

I begin by rewriting the equation y^2+2y=4x^2+3 as
y^2+2y-4x^2=3 thus allowing me to recognize that the powers or x/y ^2 are opposite, and thus the type of conic section is a hyperbola.
I complete the square for the y(s) of the equation which gives me (y+1)^2-1
and I now rewrite the equation as (y+1)^2-4x^2=3+1, (y+1)^2-4x^2=4
there is no way of completing the square for 4x^2 however since I know the equation is in the form (x-h)^2 I will rewrite -4x^2 as -4(x-0)^2
I will rewrite the equation again as:
(y+1)^2-4(x-0)^2=4 and again in the form of (y-k)^2/a^2-(x-h)^2/b^2=1
as, (y+1)^2/4 - (x-0)^2/1 = 1
thus I have a^2=4, b^2=1, and a=2, b=1
I know have all the necessary things to solve for the question.
Center = (h,k)= (0,-1)
vertices are ± a from center on axis, a =2
thus vertices are (0,-1 ± 2 )
foci are ± c from center, c^2=a^2+b^2= 1+4= √5
foci: (0, -1±√5)

That is all my work/solutions for this problem.
Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.

 

[HallsofIvy]

Homework Statement

Identify the type of conic section whose equation is given as y^2+2y=4x^2+3
Also find its vertices and foci
(y-k)^2/a^2-(x-h)^2/b^2=1

Homework Equations

I believe when you have x&y ^2 in an equation it can be either an ellipse or a hyperbola but if the signs are opposite it is a hyperbola. vertices ± a from center, foci ± c from center.

The Attempt at a Solution

I begin by rewriting the equation y^2+2y=4x^2+3 as
y^2+2y-4x^2=3 thus allowing me to recognize that the powers or x/y ^2 are opposite, and thus the type of conic section is a hyperbola.
I complete the square for the y(s) of the equation which gives me (y+1)^2-1
and I now rewrite the equation as (y+1)^2-4x^2=3+1, (y+1)^2-4x^2=4
there is no way of completing the square for 4x^2 however since I know the equation is in the form (x-h)^2 I will rewrite -4x^2 as -4(x-0)^2

Yes, that is "completing the square" for $x^2$.

I will rewrite the equation again as:
(y+1)^2-4(x-0)^2=4 and again in the form of (y-k)^2/a^2-(x-h)^2/b^2=1
as, (y+1)^2/4 - (x-0)^2/1 = 1
thus I have a^2=4, b^2=1, and a=2, b=1
I know have all the necessary things to solve for the question.
Center = (h,k)= (0,-1)
vertices are ± a from center on axis, a =2
thus vertices are (0,-1 ± 2 )
foci are ± c from center, c^2=a^2+b^2= 1+4= √5
foci: (0, -1±√5)

That is all my work/solutions for this problem.
Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.

It all looks correct to me.