Sketch x^2 +4y^2 + 6x -8y +9 = 0.

Question

Since the x squared and the y squared terms are both positive we know that this is the equation of an ellipse. We will need to complete the square on both the x and y terms. This will let us identify the center and the length of the major and minor axis.

Answer 1

x^2 +6x +4y^2 -8y = -9

x^2+6x +(6/2)^2 + 4(y^2 -2 +(2/2)^2) = -9 +9 +4 = 4

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

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

(x-h)^2)/a^2 + (y-k)^2)/b^2 = 1

(h,k) = center = (-3,1), a and b are semi axes = 2 and 1

2 = horizontal semi axis, 1 = vertical semi axis

vertices = (-1,1) (-5,1), (-3,2), (-3,0)

the ellipse is tangent to the x axis and otherwise all within quadrant II

Answer 2

x² + 4y² + 6x - 8y + 9 = 0

x² + 6x + 4y² - 8y = -9

(x² + 6x + ___) + 4(y² - 2y + ____) = -9

(x² + 6x + 9) + 4(y² - 2y +1) = -9 + 9 + 4 ←added "9" to both sides to complete the square for "x"; added 4 to both sides to complete the square for "y"

(x + 3)² + 4(y - 1)² = 4

(x + 3)²/4 + (y - 1)² = 1 ←divided everything by "4"

This is an ellipse centered at (-3, 1) with vertices at (-5, 1) and (-1, 1) and co-vertices at (-3, 2) and (-3, 0)

Answer 3

If you remember how to complete the square, this is like completing two squares at the same time (once for x and once for y).

Rearrange the equation by putting your x's together and y's together, and move the constant to the other side of the equal sign.

x² + 6x + 4y² - 8y = -9

Use (^b/₂)² for the x and y portions, note I factored 4 out from the y's first. This also means that when we get +1 for completing y's square, it actually adds 4 to the other side of the equation because of the parenthesis.

x² + 6x + 9 + 4(y² - 2y + 1) = -9 + 9 + 4

Factor.

(x + 3)² + 4(y - 1)² = 4

Divide everything by 4 (ellipse formula wants everything to equal 1).

(x + 3)²/4 + (y - 1)² = 1

Convert to ellipse formula (4 is a² in this case so a = 2).

(x + 3)²/2² + (y - 1)² = 1

This will create an ellipse centered at (-3, 1) with a major (horizontal) radius of 2 and a minor (vertical) radius of 1.

Answer 4

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Visit the above graph and slide point P to see that the sum of the distances from P to each focus is a constant (2a).