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Difficulty: MediumConic Sections

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+4y2+6x−8y+9=0x^2 + 4y^2 + 6x - 8y + 9 = 0. What is the length of the major axis of this ellipse?

  1. A
    1
  2. B
    2
  3. 4Answer
  4. D
    8
  5. E
    16

Answer

4
To find the length of the major axis, rewrite the general equation of the ellipse in standard form by completing the square. Grouping the terms yields (x2+6x)+4(y2−2y)=−9(x^2 + 6x) + 4(y^2 - 2y) = -9. Completing the square for both variables gives (x+3)2−9+4[(y−1)2−1]=−9(x+3)^2 - 9 + 4[(y-1)^2 - 1] = -9, which simplifies to (x+3)2+4(y−1)2=4(x+3)^2 + 4(y-1)^2 = 4. Dividing both sides by 4 gives the standard form (x+3)24+(y−1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1. In this form, the horizontal axis is the major axis because the denominator under the xx-term (a2=4a^2 = 4) is larger than the denominator under the yy-term (b2=1b^2 = 1). Since a2=4a^2 = 4, the semi-major axis is a=2a = 2. Therefore, the total length of the major axis is 2a=2(2)=42a = 2(2) = 4.

Step-by-Step Solution

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
(x2+6x)+(4y2−8y)=−9(x^2 + 6x) + (4y^2 - 8y) = -9
Grouping like terms allows completing the square for each variable independently.
2
Factor out the coefficient of y2y^2 from the yy-terms.
(x2+6x)+4(y2−2y)=−9(x^2 + 6x) + 4(y^2 - 2y) = -9
Before completing the square, the leading coefficient of the squared terms inside the parentheses must be 1.
3
Complete the square for both the xx and yy expressions by adding and subtracting the square of half of the linear coefficients.
((x+3)2−9)+4((y−1)2−1)=−9((x+3)^2 - 9) + 4((y-1)^2 - 1) = -9
This rewrites the quadratic expressions into perfect square trinomial form.
4
Distribute the coefficients and simplify the constant terms.
(x+3)2−9+4(y−1)2−4=−9  ⟹  (x+3)2+4(y−1)2−13=−9  ⟹  (x+3)2+4(y−1)2=4(x+3)^2 - 9 + 4(y-1)^2 - 4 = -9 \implies (x+3)^2 + 4(y-1)^2 - 13 = -9 \implies (x+3)^2 + 4(y-1)^2 = 4
Isolating the squared terms on one side helps convert the equation to the standard form of an ellipse.
5
Divide both sides of the equation by 4 to set the right side equal to 1.
(x+3)24+(y−1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1
The standard form of a horizontal ellipse is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.
6
Identify the values of a2a^2 and b2b^2 and calculate the length of the major axis.
a2=4  ⟹  a=2a^2 = 4 \implies a = 2. The major axis length is 2a=2(2)=42a = 2(2) = 4.
The length of the major axis is twice the length of the semi-major axis (aa).

Key Concept

Rewriting the general equation of an ellipse into standard form by completing the square to find its key features, such as the length of the major axis.

Alternative Method

Another way to find the length of the major axis is to find the vertices of the ellipse by finding the maximum and minimum x-values where the equation has real solutions for y, though completing the square is the standard and most direct method.
Estimated Time:1m 30s
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Topics

Conic Sections

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