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Chapter 6: Problem 48

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$y^{2}-6 y-4 x+21=0$$

Short Answer

Expert verified
The graph of the equation is a parabola.

Step by step solution

01

Identify the equation form

By referring to the general equations for a circle, a parabola, an ellipse and a hyperbola, we know that the given equation presents features corresponding to a parabola, where the only squared term is \(y^2\), while \(x\) does not have a square term.
02

Rearrange the equation into standard form

Write the equation \(y^2 - 6y - 4x + 21 = 0\) as \(y^2 - 6y = 4x - 21\). Then complete the square on the left side, giving \(y - 3)^2 = 4x - 21 + 9\), which simplifies to \((y - 3)^2 = 4x - 12\). This is in the form \((y - k)^2 = 4a(x - h)\), where \(k = 3\), \(a = 1\), and \(h = -3\).
03

Identify the curve

This is a form of a vertical parabola, opening to the right side (since the coefficient of x is positive). It is centered at \((-3, 3)\)

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Most popular questions from this chapter

\(r=6+12 \cos \theta\)

In your own words, define the term eccentricity and explain how it can be used to classify conics.

Consider the graph of \(r=f(\sin \theta)\). (a) Show that if the graph is rotated counterclockwise \(\pi / 2\) radians about the pole, the equation of the rotated graph is \(r=f(-\cos \theta)\). (b) Show that if the graph is rotated counterclockwise \(\pi\) radians about the pole, the equation of the rotated graph is \(r=f(-\sin \theta)\). (c) Show that if the graph is rotated counterclockwise \(3 \pi / 2\) radians about the pole, the equation of the rotated graph is \(r=f(\cos \theta)\).

In Exercises 73-78, solve the trigonometric equation. $$ 6 \cos x-2=1 $$

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Parabola } & (6,0) \\ \end{array} $$
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