Checking if ellipse question is correct

[Cake]

/*Solved*/ Checking if ellipse question is correct.

Hey Guys,

I am just wondering if my steps and answers look correct. I would love the feed back.

Thanks.

(3) Find the center, vertices, and foci of the ellipse having the following equation:

. . .\(\displaystyle 3x^2\, +\, 4y^2\, +\, 12x\, -\, 8y\, -\, 32\, =\, 0\)

My work:

\(\displaystyle 3x^2 + 12x + 4y^2 - 8y = 32\)

\(\displaystyle 3(x^2 + 4x + 4) 4(y^2 - 2y + 1) = 32\)

\(\displaystyle \dfrac{3(x + 2)^2}{48} + \dfrac{4(y - 1)^2}{48} = \dfrac{48}{48}\)

\(\displaystyle \dfrac{(x + 2)^2}{16} + \dfrac{(y - 1)^2}{12} = 1\)

\(\displaystyle a = \sqrt{16} = 4\)

\(\displaystyle b = \sqrt{12} = 2\sqrt{3}\)

\(\displaystyle c = \sqrt{16 - 12} = \sqrt{4} = 2\)

center: (-2, 1)
V₁: (2, 1), V₂: (6, 1)
F₁: \(\displaystyle (2 + 2\sqrt{3},\, 1)\), F₂: \(\displaystyle (2 - 2\sqrt{3},\, 1)\)

**This question has been answered**

 

[stapel]

I am just wondering if my steps and answers look correct. I would love the feed back.

(3) Find the center, vertices, and foci of the ellipse having the following equation:

. . .\(\displaystyle 3x^2\, +\, 4y^2\, +\, 12x\, -\, 8y\, -\, 32\, =\, 0\)

My work:

\(\displaystyle 3x^2 + 12x + 4y^2 - 8y = 32\)

\(\displaystyle 3(x^2 + 4x + 4) 4(y^2 - 2y + 1) = 32\)

Note: Be careful with the signs and numbers! There should still be a "plus" between the x-terms and the y-terms, and the values you've added on the left-hand side should also be added to the right-hand side, in order to preserve the equality. Many instructors will stop grading the exercise at the above line, marking everything that follows as incorrect.

\(\displaystyle \dfrac{3(x + 2)^2}{48} + \dfrac{4(y - 1)^2}{48} = \dfrac{48}{48}\)

Here, the "plus" magically reappears, and the right-hand side has changed, for no apparent reason, from "32" to "48". Many instructors, if they hadn't stopped at the previous line, would stop here.

Forewarned is fore-armed!