Question

X^2+6y^2+6x+36y=-27

Identify the center, the vertices and foci of the ellipse

You should have 4 vertices, and 2 foci's

Answer 1

`\textsf{Center:}\quad (0,0)`

`\textsf{Vertices:}\quad (-9,-3)\;\textsf{and}\;(3,-3)`

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

`\textsf{Foci:}\quad \left(-3-√(30),-3\right)\;\textsf{and}\;\left(-3+√(30),-3\right)`

Explanation:

Given equation of the ellipse:

`x^2+6y^2+6x+36y=-27`

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

Collect like terms:

`x^2+6x+6y^2+36y=-27`

Factor out 6 from the terms in y:

`x^2+6x+6(y^2+6y)=-27`

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+6\left(y^2+6y+\left((6)/(2)\right)^2\right)=-27+\left((6)/(2)\right)^2+6\left((6)/(2)\right)^2`

Simplify:

`x^2+6x+9+6\left(y^2+6y+9\right)=-27+9+54`

`x^2+6x+9+6\left(y^2+6y+9\right)=36`

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

`(x+3)^2+6(y+3)^2=36`

Divide both sides of the equation by 36:

`((x+3)^2)/(36)+(6(y+3)^2)/(36)=(36)/(36)`

`((x+3)^2)/(36)+((y+3)^2)/(6)=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 = -3`
• 
  `a^2 = 36\implies a=√(36)=6`
• 
  `b^2 = 6 \implies b=√(6)`

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

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)\\&=(-3\pm 6, -3)\\&=(-9,-3)\;\textsf{and}\;(3,-3)\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,-3\pm √(6)\right)\\&=\left(-3,-3-√(6)\right)\;\textsf{and}\;\left(-3,-3+√(6)\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&=√(36-6)\\c&=√(30)\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\pm √(30),-3\right)\\&=\left(-3-√(30),-3\right)\;\textsf{and}\;\left(-3+√(30),-3\right)\end{aligned}`