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Chapter 10: Problem 57

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$9 x^{2}+4 y^{2}-18 x+16 y-119=0$$

Short Answer

Expert verified
The graph of the given equation represents an ellipse with a vertical major axis.

Step by step solution

01

Rearrange terms

Rearrange the terms in the given equation so that all terms involving x are grouped together, and all terms involving y are grouped together: \(9x^{2}-18x + 4y^{2}+16y = 119\).
02

Complete the square

Complete the square for the terms involving x and those involving y so that the equation can be written in a standard form. To complete the squares, take half of the coefficient of x (or y) in the linear term (x or y term), square it, and add it to both sides of the equation. This will give \(9(x^{2}-2x+1) + 4(y^{2}+4y+4) = 119 + 9*1 + 4*4\). Simplify this to get \(9(x-1)^{2} + 4(y+2)^{2} = 143\).
03

Write in standard form

Divide all terms by 143 to obtain the equation in standard form, which expresses the equation as a sum of fractions: \[(x-1)^{2}/(143/9) + (y+2)^{2}/(143/4) = 1.\] Simplify this to get the standard form of the equation: \[(x-1)^{2}/(143/9) + (y+2)^{2}/(143/4) = 1.\]
04

Identify the conic section

Because both squared variables are added together and their coefficients are different, the graph represents an ellipse. The larger denominator under the squared terms is under (y+2)^{2}, so the major axis is vertical.

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