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Chapter 6: Problem 68

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 x^{2}-y^{2}+4 x+2 y-1=0$$

Short Answer

Expert verified
The given equation corresponds to the form of a hyperbola.

Step by step solution

01

Observe the given equation

The equation provided in the exercise is of the form \(Ax^{2}+By^{2}+Cx+Dy+E=0\). Upon observation, it is noticeable that both x and y are squared.
02

Rearrange the equation to simplify

We rearrange the equation into groups of x's and y's separately to simplify it. \(4x^{2}+4x-y^{2}+2y-1=0\). The equation can then be expressed in the form \((2x)^{2}+2*2x-(y)^{2}+2y-1=0\).
03

Modify into standard form of conic section

Now, modify the equation into one of the standard forms of conic sections. Notice that if we make the coefficients of \(x^{2}\) and \(y^{2}\) as 1, we get a hyperbola form \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2} = 1\).
04

Compare the equation obtained with hyperbola standard form

The rearranged equation from step 3 is comparable to the standard form of a hyperbola where a=2 and b=1. Coefficients of \(x^{2}\) and \(y^{2}\) in the equation from step 3 have different signs, which makes it look like a hyperbola.

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