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Chapter 10: Problem 90

What happens to the shape of the graph of \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) as \(\frac{c}{a} \rightarrow 0,\) where \(c^{2}=a^{2}-b^{2} ?\)

Short Answer

Expert verified
As \(\frac{c}{a} \rightarrow 0,\) the equation of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) becomes a circle of radius \(a\), represented by the equation \(x^{2} + y^{2} = a^{2}\).

Step by step solution

01

Define and understand the equations

Start with the given equation for an ellipse: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the relationship between \(a, b, c,\) where \(c^{2}=a^{2}-b^{2}.\)
02

Determine what happens as \(\frac{c}{a}\) approaches 0

As \(c\) approaches 0, from the equation \(c^{2}=a^{2}-b^{2},\) it follows that \(a^{2}-b^{2}\) also approaches 0. That means \(a^{2}\) approaches \(b^{2}\), or in other words, \(a\) approaches \(b\).
03

Substitute into the equation of the ellipse

When substituting \(a = b\) into the equation of the ellipse, it transforms to \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}=1\), which simplifies to \(x^{2} + y^{2} = a^{2}.\) This equation represents a circle of radius \(a\).

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