Problem 55 - Precalculus

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 x^{2}-y^{2}-4 x-3=0$$

Step-by-Step Solution

Answer

The graph of the given equation is a hyperbola.

Step 1: Rewrite the equation in standard form

Rewrite the equation into a form where the squared terms are together, so you can identify the standard form: \( 4x^{2}-4x-y^{2}=3 \). Completing the square on the \( x \) terms, which includes half of the coefficient of the \( x \) term (which is -2), squared and added to both sides gives: \( 4(x-1)^{2}-y^{2}=7 \). Divide every term by 7 to get: \( (x-1)^{2}-\frac{y^{2}}{7}=1 \).

Step 2: Identify the equation's type

Compare the obtained equation with the standard form equations for a circle, a parabola, an ellipse or a hyperbola. Here, this equation can be matched to the standard form of a hyperbola: \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \), where \( h=1, a^{2}=1, k=0, b^{2}=7 \).

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola is essential for recognizing and distinguishing it from other conic sections. It is expressed as \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \) or \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), where \( (h, k) \) is the center of the hyperbola, \( a \) is the distance from the center to a vertex along the axis of symmetry, and \( b \) is the distance from the center to the co-vertex along the conjugate axis.

To identify a hyperbola, look for a negative sign between the two variables' squared terms in the equation. Each variable is part of a fraction that equals 1, with contrasting signs indicating the hyperbola's two branches are mirror images across the axis. For the given equation \( 4x^{2}-y^{2}-4x-3=0 \) when rewritten in standard form \( (x-1)^{2}-\frac{y^{2}}{7}=1 \) confirms it as a hyperbola.

Completing the Square

Completing the square is a technique to transform a quadratic expression into a perfect square trinomial. This method is crucial when dealing with conic sections, as it allows us to rewrite equations in their standard forms. The process involves the following steps:

• Isolate the variable terms from the constant terms.
• If the quadratic coefficient (the coefficient of the squared term) is not 1, factor it out from both of the variable terms.
• Take half the coefficient of the linear term (the term with just the variable), square it, and add it to both sides of the equation to balance it.
• Rewrite the quadratic expression as the square of a binomial.

For instance, the equation \( 4x^{2}-4x-y^{2}=3 \) can be rearranged by completing the square to \( 4(x-1)^{2} - y^{2}=7 \) in the solution provided.

Conic Sections in Precalculus

Conic sections are the curves obtained by intersecting a cone with a plane. In precalculus, students learn about four primary types of conic sections: circles, parabolas, ellipses, and hyperbolas. These shapes come with distinct standard form equations, which help in identifying and analyzing their properties. Here’s a quick overview:

• Circle: \( (x-h)^{2} + (y-k)^{2} = r^{2} \) where \( r \) is the radius and \( (h, k) \) is the center.
• Parabola: \( (y-k) = a(x-h)^{2} \) or \( (x-h) = a(y-k)^{2} \) where the vertex is \( (h, k) \).
• Ellipse: \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \) with \( a \) and \( b \) being the lengths of the major and minor axes, respectively.
• Hyperbola: \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \) with two opposed opening parabolic branches.

Understanding these forms is pivotal for students as they solve conic sections problems and analyze the graphs of these equations. The exercise provided begins with an equation that does not immediately reveal its conic nature. Using the strategy of completing the square and rewriting the equation leads to identification as a hyperbola.