Ellipse: Standard Equation

Step 1: Determine the following:

➢ The coordinates of the center (h, k).

➢ The distance of half the minor axis (b).

➢ The length of half the major axis (a).

➢ The orientation of the major axis.

Center: Since the foci are equidistant from the center of the ellipse the center can be determine by finding the midpoint of the foci.

$\left(h, k\right)=\left(\frac{2+\left(-4\right)}{2}, \frac{1+1}{2}\right)=\left(-\frac{2}{2},\frac{2}{2}\right)=\left(-1,1\right)$

Length of b: The minor axis is given as 10, which is equal to 2b.

$2b=10\to b=5$

Length of a: To find a the equation c² = a² + b² can be used but the value of c must be determined. Since c is the distance from the foci to the center, take either foci and determine the distance to the center. Then solve for a.

Foci (2, 1): $c=\left|2-\left(-1\right)\right|=\left|3\right|=3$

Foci (-4, 1): $c=\left|-4- \left(-1\right)\right|=\left|-3\right|=3$

c = 3

$c^2=a^2-b^2\to a=\sqrt{c^2+b^2}$

$a=\sqrt{3^2+5^2}\to a=\sqrt{34}$

Orientation of major axis: Since the two foci fall on the horizontal line y = 1, the major axis is horizontal.

Step 2: Substitute the values for h, k, a and b into the equation for an ellipse with a horizontal major axis.

Horizontal major axis equation:

$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Substitute values:

$\frac{{\left[x-\left(-1\right)\right]}^2}{{\sqrt{34}}^2}+\frac{{\left(y-1\right)}^2}{5^2}=1$

Simplify:

$\frac{{\left(x+1\right)}^2}{{\sqrt{34}}^2}+\frac{{\left(y-1\right)}^2}{5^2}=1$

Step 1: Group the x- and y-terms on the left-hand side of the equation.

x² + 3y² - 4x - 18y + 4 = 0

$(x^2-4x)+(3y^2-18y)+4=0$

Step 2: Move the constant term to the right-hand side.

$(x^2-4x)+(3y^2-18y)=-4$

Step 3: Complete the square for the x- and y-groups.

$(x^2-4x)+(3y^2-18y)=-4$

Complete the square for the x-group

(x² - 4x)

Take the coefficient of the x-term, divide by 2 and square the result.

$\frac{-4}{2}=-2;{\left(-2\right)}^2=4$

Add the result to the x-group.

(x² - 4x + 4)

Complete the square for the y-group

(3y² - 18y)

Factor out a 3 so they y²-coefficient is 1

3(y² - 6y)

Take the coefficient of the y-term, divide by 2 and square the result.

$\frac{-6}{2}=-3;{\left(-3\right)}^2=9$

Add the result to the y-group.

3(y² - 6y + 9)

Final result.

$(x^2-4x+4)+3(y^2-6y+9)=-4+?$

Step 4: Add whatever was added to the left-hand side to the right-hand side.

$(x^2-4x+4)+3(y^2-6y+9)=-4+4+3\left(9\right)$

$(x^2-4x+4)+3(y^2-6y+9)=27$

Step 5: Write the x-group and y-group as perfect squares.

${\left(x-2\right)}^2+3{\left(y-3\right)}^2=27$

Step 6: Divide both sides by the value on the right-hand side.

$\frac{{\left(x-2\right)}^2}{27}+\frac{3{\left(y-3\right)}^2}{27}=1$

$\frac{{\left(x-2\right)}^2}{27}+\frac{{\left(y-3\right)}^2}{9}=1$