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Chapter 6: Problem 61

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 x^{2}+3 y^{2}+8 x-24 y+51=0$$

Short Answer

Expert verified
The graph of the given equation represents an ellipse.

Step by step solution

01

Organizing the Given Equation

Firstly, group together the x terms and y terms, which gives \(4x^{2}+8x+3y^{2}-24y = -51\).
02

Completing the Square for x and y

To make the x terms look like a square, rewrite \(4x^{2}+8x\) as \(4(x^{2}+2x)\). For completing the square, add square of half the coefficient of x in the square bracket which is \((2/2)^{2}=1\), now add 4 on both sides. The x terms become \(4(x+1)^{2}\). Do the same for y terms, which gives \(3(y-4)^{2}\). Hence our new equation becomes \(4(x+1)^{2} + 3(y-4)^{2} = -51+4+48\)
03

Simplification and Identification

The simplified equation is \(4(x+1)^{2} + 3(y-4)^{2} = 1\). Divide entire equation by 1, this doesn't change the equation: \((x+1)^{2}/(1/4) + (y-4)^{2}/(1/3) = 1\). This equation is the standard form of an ellipse.

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