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Chapter 6: Problem 66

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$9 x^{2}+4 y^{2}-90 x+8 y+228=0$$

Short Answer

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

Step by step solution

01

Rewrite the equation

First, we need to rewrite the given equation in the form where the terms comprising \(x\) and \(y\) are grouped separately: the equation \(9x^{2}-90x+4y^{2}+8y=-228\) .
02

Complete the square for x terms

We should group the terms involving \(x\) and complete the square. In this case it is \(9x^{2} - 90x\). Half of the value of the coefficient of \(x\), squared is added to both sides of the equation to complete the square. This would be \((90/2*9)^{2} = 25\), leading to \(9(x^{2}-10x+25)=5^2\). Simplifying this gives \(9(x-5)^{2}=25\).
03

Complete the square for y terms

We do the same for the \(y\) part of the equation \(4y^{2} + 8y\), adding \((8/2*4)^{2} = 1\), which gives us \(4(y^{2}+2y+1)=1^2\). Simplifying this, we get \(4(y+1)^{2}=1\).
04

Final equation

We combine the results of step 2 and step 3 to get the equation in standard form. The final equation is \(9(x-5)^{2}+4(y+1)^{2}=0\).
05

Identify the type of conic

The standard form \(a(x-h)^{2}+b(y-k)^{2}=0\), where \(a\) and \(b\) are positive and not equal to each other is that of an ellipse.

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