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Chapter 6: Problem 56

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 given equation is a parabola.

Step by step solution

01

Identify the Form

To classify the given equation, it's important to note that the general form for a circle is \( (x-a)^{2}+(y-b)^{2}=r^{2} \), the form for a standard parabola can be \( y=a(x-h)^{2}+k \) or \( x=a(y-k)^{2}+h \), the form for ellipse is \( \frac{(x-h)^{2}}{a^{2}}+ \frac{(y-k)^{2}}{b^{2}}=1 \) and the form for a hyperbola is \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \) or \( \frac{(y-k)^{2}}{b^{2}}-\frac{(x-h)^{2}}{a^{2}}=1 \)
02

Analyze the Given Equation

Rewrite the given equation \( y^{2}-6y-4x+21=0 \) as \( y^{2}-6y=-4x+21 \) or \( y^{2}-6y+9=-4x+30 \). This equation now looks similar to \( y^{2}-2hy=-4x+k \) which resembles the standard form of a parabola \( y=a(x-h)^{2}+k \).
03

Equate to Zero

To get the same form, complete the square on the left hand side of the equation, \( (y-3)^{2}=30-4x \), then rewrite the equation as \( (y-3)^{2}=4(7.5-x) \). Now, it's clear that this is in the form of a parabola, where a=4, h=7.5 and k=3.

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