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Chapter 6: Problem 57

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 equation \(y^{2}-6 y-4 x+21=0\) represents a parabola.

Step by step solution

01

Rewrite the equation

First, write given quadratic equation in standard form: \(y^{2}-6y+9 = 4x -12 + 9\). Here, we have added 9 on both sides to complete the square for y term.
02

Simplify the equation

Simplify the equation to get it in the standard form of one of the conic section equations: \((y-3)^2 = 4(x - 3)\)
03

Identify the conic section

This equation is in the form of \( (y-K)^2 = 4a(x-H)\) where 'a' is a constant, and 'H' and 'K' are the coordinates at the vertex. This form is the standard form of a parabola that opens either upwards or downwards. In this case, it opens to the right, which is confirmed by 'x' being by itself on one side of the equation. Therefore, the given equation is a parabola.

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