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Chapter 10: Problem 25

Classify each of the following as the equation of either a circle, an ellipse, a parabola, or a hyperbola. $$ x^{2}+y^{2}-6 x+10 y-40=0 $$

Short Answer

Expert verified
Circle

Step by step solution

01

Identify the Type of Conic Section

Compare the given equation to the general forms of a circle, ellipse, parabola, and hyperbola. Notice that both x and y are squared in the equation.
02

Complete the Square for x and y

To classify the conic section, rewrite the equation by completing the square. Start by grouping the x terms and the y terms together: \[ x^2 - 6x + y^2 + 10y = 40 \]
03

Complete the Square for x

Add and subtract the same value to complete the square for the x terms: \[ x^2 - 6x + 9 - 9 \] This simplifies to \[ (x-3)^2 - 9 \]
04

Complete the Square for y

Add and subtract the same value to complete the square for the y terms: \[ y^2 + 10y + 25 - 25 \] This simplifies to \[ (y+5)^2 - 25 \]
05

Substitute Completed Squares Back into the Equation

Substitute these modified terms back into the equation to obtain: \[ (x-3)^2 - 9 + (y+5)^2 - 25 = 40 \]
06

Simplify the Equation

Combine and simplify all the constants on one side: \[ (x-3)^2 + (y+5)^2 - 34 = 40 \] \[ (x-3)^2 + (y+5)^2 = 74 \]
07

Identify the Resulting Conic Section

The equation now matches the general form of a circle: \[ (x-h)^2 + (y-k)^2 = r^2 \] with center \((h, k) = (3, -5)\) and radius \(r = \sqrt{74}\). Hence, it is a circle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of a Circle
Understanding the equation of a circle is crucial when working with conic sections. The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\). Here, \(h\) and \(k\) represent the coordinates of the circle's center, and \(r\) represents the radius.
For instance, if you come across an equation like \((x - 3)^2 + (y + 5)^2 = 74\), you can identify it as a circle with:
  • Center at \( (3, -5) \)
  • Radius \( \sqrt{74} \)

Notice how the equation is structured:
  • Each variable, \(x\) and \(y\), is squared.
  • There are no additional x or y terms outside the parentheses.
  • All terms are on one side of the equation, with a constant on the other side.
Recognizing these characteristics helps quickly identify circle equations.
Completing the Square
Completing the square is an important algebraic technique used to convert quadratic equations into a more recognisable form. Here's how it works:
  • Suppose we have \ x^2 - 6x \. To complete the square, follow these steps:
1. Take half of the coefficient of \(x\) (which is -6), square it to get \(-3)^2 = 9\).
2. Add and subtract this square inside the equation to form a perfect square trinomial: \( x^2 - 6x + 9 - 9 = (x-3)^2 - 9 \).
  • Do the same for the \(y\) terms. Adding and subtracting \(25\), derived similarly, converts \ y^2 + 10y \ to \( (y+5)^2 - 25 \).
Once we apply these changes to the given equation, it transforms into completed squares, making it easier to identify its conic section form later on.
Identifying Conic Sections
When you are given a quadratic equation in two variables, identifying the type of conic section it represents is essential. Conic sections (circle, ellipse, parabola, hyperbola) can be recognized by the following characteristics:
  • A parabola has only one squared term, either \(x^2\) or \(y^2\), but not both.
  • An ellipse and a circle both have \(x^2\) and \(y^2\) with the same sign, but a circle's coefficients of the squared terms are equal.
  • A hyperbola has \(x^2\) and \(y^2\) with opposite signs.

For the equation \ x^2 + y^2 - 6x + 10y - 40 = 0 \, we see both \(x^2\) and \(y^2\) terms with the same coefficients (which are both 1 here):
- By completing the square, it simplifies the equation into the form of a circle,
- Through these steps, each equation can be classified into its respective conic section. So, knowing the general forms and applying techniques like completing the square is key.

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Most popular questions from this chapter

The standard form for equations of horizontal or vertical hyperbolas centered at \((h, k)\) are as follows: $$ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 $$ (Graph can't copy) $$ \frac{(y-k)^{2}}{b^{2}}-\frac{(x-h)^{2}}{a^{2}}=1 $$ The vertices are as labeled and the asymptotes are $$ y-k=\frac{b}{a}(x-h) \text { and } y-k=-\frac{b}{a}(x-h) $$ For each of the following equations of hyperbolas, complete the square, if necessary, and write in standard form. Find the center, the vertices, and the asymptotes. Then graph the hyperbola. $$ 4 y^{2}-25 x^{2}-8 y-100 x-196=0 $$

Find an equation of a circle satisfying the given conditions. Center \((3,-5)\) and tangent to (touching at one point \()\) the \(y\) -axis

On a piece of graph paper, draw a line and a point not on the line. Then plot several points that are equidistant from the point and the line. What shape do the points appear to form? How could you confirm this?

The standard form for equations of horizontal or vertical hyperbolas centered at \((h, k)\) are as follows: $$ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 $$ (Graph can't copy) $$ \frac{(y-k)^{2}}{b^{2}}-\frac{(x-h)^{2}}{a^{2}}=1 $$ The vertices are as labeled and the asymptotes are $$ y-k=\frac{b}{a}(x-h) \text { and } y-k=-\frac{b}{a}(x-h) $$ For each of the following equations of hyperbolas, complete the square, if necessary, and write in standard form. Find the center, the vertices, and the asymptotes. Then graph the hyperbola. $$ \frac{(x-5)^{2}}{36}-\frac{(y-2)^{2}}{25}=1 $$

The equation \(x^{2}+y^{2}=\frac{81}{4},\) where \(x\) and \(y\) represent the number of meters from the center, can be used to draw the outer circle on a wrestling mat used in International, Olympic, and World Championship wrestling. The equation \(x^{2}+y^{2}=16\) can be used to draw the inner edge of the red zone. Find the area of the red zone. (IMAGE CANNOT COPY)
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