Problem 64 - Precalculus

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 y^{2}+4 x^{2}-24 x+35=0$$

Step-by-Step Solution

Answer

The graph of the equation \(4 y^{2} + 4 x^{2} - 24 x + 35 = 0\) is a circle.

Step 1: Normalize the equation

Initially, rearrange the given equation to identify the standard form of the conic section. In this instance, normalize the given equation by grouping the terms with common variables and factoring out the coefficients of the squares. So, the equation becomes \(4(x^{2} - 6x) + 4y^{2}= -35\).

Step 2: Complete the square

To write this equation in a standard form, complete the square for the \(x\) terms. To complete the square, take half the coefficient of \(x\), square it and add to both sides. For \(x^{2} - 6x\), half of 6 is 3, and 3 squared is 9. Therefore, add 9 to both sides to complete the square. The equation now becomes \(4[(x^{2}-6x+9) - 9] + 4y^{2} = -35 + 36. Simplify the equation to get \(4[(x-3)^{2}] + 4y^{2} = 1\).

Step 3: Identify the conic section

The normal form for the equation of a circle is \((x-h)^{2} + (y-k)^{2} = r^{2}\) which can be written as \((x-h)^{2}/r^{2} + (y-k)^{2}/r^{2} = 1\), where \( (h, k) \) are the coordinates of the center of the circle, and \( r \) is the radius. Now, dividing our equation from step 2 by 4 gives \((x-3)^{2} + y^{2} = 1\), which is in the form of a circle, where the center of the circle is (3, 0) and radius is 1.

Key Concepts

Circle

A circle is a special type of conic section. It is the set of all points in a plane that are equidistant from a fixed point, known as the center. This distance is called the radius. Circles are characterized by their symmetrical shape and are commonly found in everyday life—from wheels to clocks.

The standard form of a circle's equation is

• displayed as: \((x-h)^2 + (y-k)^2 = r^2\),

where

• \((h, k)\) represents the center, and
• \(r\) is the radius.

This format helps us quickly identify the circle's important attributes, such as where it is located on the graph and how big it is. Understanding the equation of a circle is key to classifying conic sections accurately.

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This process is essential when working with conic sections, especially when converting an equation to its standard form.

The method involves these steps:

• Start with the quadratic expression, typically \(x^2 + bx\).
• Find half of the coefficient of \(x\), which is \(b/2\).
• Square this half, giving \((b/2)^2\).
• Add and subtract this square from the expression, forming a perfect square trinomial.

This activity ultimately rewrites the expression as

• a binomial square: \((x + b/2)^2 - (b/2)^2\).

In conic sections, especially circles, completing the square allows for easy identification of the center and radius. It's a versatile tool for rewriting equations and clarifying their geometric properties.

Standard Form of Conics

The standard form of conic sections—circles, ellipses, parabolas, and hyperbolas—allows us to easily recognize and analyze these shapes within a coordinate plane. Each conic section has its unique standard representation, helping to set them apart and understand their geometric characteristics.

Common forms include:

• For circles: \((x-h)^2 + (y-k)^2 = r^2\), indicating a precise center and radius.
• For ellipses: \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\), defining the lengths of axes and the center.
• For parabolas: \(y-k = a(x-h)^2\) or \(x-h = a(y-k)^2\), highlighting the vertex and direction of opening.
• For hyperbolas: \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), showing asymptotes, center, and axes.

Understanding these forms helps in easily identifying the type and properties of the conic based on its equation. It creates a straightforward path from formula to graph.