Question

Identify the center, the vertices and the foci of the ellipse

x^2+3y^2-6x+6y+6=0

Answer 1

`\textsf{Center:}\quad (3,-1)`

`\textsf{Vertices:}\quad \left(3-√(6), -1\right)\;\textsf{and}\;\left(3+√(6), -1\right)`

`\textsf{Co-vertices:}\quad \left(3,-1-√(2)\right)\;\textsf{and}\;\left(3,-1+√(2)\right)`

`\textsf{Foci:}\quad \left(1,-1\right)\;\textsf{and}\;\left(5,-1\right)`

Explanation:

Given equation of the ellipse:

`x^2+3y^2-6x+6y+6=0`

Rewrite the given equation into the standard form of an ellipse by completing the square for both the x and y terms.

Move the constant to the right side of the equation and collect like terms on the left side of the equation:

`x^2-6x+3y^2+6y=-6`

Factor out the coefficient of the y² term from the terms in y:

`x^2-6x+3(y^2+2y)=-6`

Add the square of half the coefficient of the term in x to both sides of the equation. Add the square of half the coefficient of the term in y inside the parentheses on the left side of the equation, and add its distributed value to the right side of the equation:

`x^2-6x+\left((-6)/(2)\right)^2+3\left(y^2+2y+\left((2)/(2)\right)^2\right)=-6+\left((-6)/(2)\right)^2+3\left((2)/(2)\right)^2`

Simplify:

`x^2-6x+9+3(y^2+2y+1)=-6+9+3`

`x^2-6x+9+3(y^2+2y+1)=6`

Factor the perfect square trinomials on the left side of the equation:

`(x-3)^2+3(y+1)^2=6`

Divide both sides of the equation by 6:

`((x-3)^2)/(6)+(3(y+1)^2)/(6)=(6)/(6)`

`((x-3)^2)/(6)+((y+1)^2)/(2)=1`

As the denominator of the x² term is larger than the denominator of the y² term, it indicates that the ellipse has a horizontal major axis, making it a horizontal ellipse.

The standard equation of a horizontal ellipse is:

`\boxed{\begin{minipage}{9.2 cm}\underline{Standard equation of a horizontal ellipse}\\\\$((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center\\ \phantom{ww}$\bullet$ $(h\pm a,k)$ are the vertices\\ \phantom{ww}$\bullet$ $(h,k\pm b)$ are the co-vertices\\ \phantom{ww}$\bullet$ $(h\pm c,k)$ are the foci where $c^2=a^2-b^2.$\end{minipage}}`

Comparing the given equation with the standard equation, we have:

• 
  `h = 3`
• 
  `k = -1`
• 
  `a^2 = 6\implies a=√(6)`
• 
  `b^2 = 2 \implies b=√(2)`

Therefore, the center (h, k) is located at (3, -1).

To find the coordinates of the vertices, substitute the values of h, k and a into the vertices formula:

`\begin{aligned}\textsf{Vertices}&=(h\pm a, k)\\&=\left(3\pm √(6), -1\right)\\&=\left(3-√(6), -1\right)\;\textsf{and}\;\left(3+√(6), -1\right)\end{aligned}`

To find the coordinates of the co-vertices, substitute the values of h, k and b into the co-vertices formula:

`\begin{aligned}\textsf{Co-vertices}&=(h, k\pm b)\\&=\left(3,-1\pm √(2)\right)\\&=\left(3,-1-√(2)\right)\;\textsf{and}\;\left(3,-1+√(2)\right)\end{aligned}`

To find the foci, we first need to find the value of c by substituting the values of a² and b² into the equation c² = a² - b²:

`\begin{aligned}c^2&=a^2-b^2\\c&=√(a^2-b^2)\\c&=√(6-2)\\c&=√(4)\\c&=2\end{aligned}`

Finally, to determine the foci, substitute the values of h, k and c into the foci formula:

`\begin{aligned}\sf Foci&=(h \pm c,k)\\&=\left(3\pm2,-1\right)\\&=\left(1,-1\right)\;\textsf{and}\;\left(5,-1\right)\end{aligned}`