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Chapter 6: Problem 53

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 given graph of the equation is an ellipse.

Step by step solution

01

Identify the variables

In this equation the variables are \(x\) and \(y\) and coefficients are with \({x^2}, {y^2}, x, y\) and the constant.
02

Rearranging the equation

Rearrange the equation by grouping the x terms and y terms together as follows: \((9x^2-18x) + (4y^2+16y) = 119\)
03

Complete the squares

In order to identify the form of the equation, we try to complete the squares for \(x\) and \(y\)After doing so, we get \((3x - 3)^2 + (2y + 4)^2 = 121\)
04

Identify the Conic Section

The equation is now in standard form for a conic section, and it can be recognized as the equation of an ellipse because it has an \(x^2\) term, a \(y^2\) term (both with positive coefficients) and an equal sign. The general form is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\), where \((h, k)\) is the center of the ellipse, and \(a\) and \(b\) are the semi-major and semi-minor axis lengths respectively.

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