Problem 34 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ x^{2}-8 y+y^{2}+11=0 $$

Step-by-Step Solution

Answer

The equation represents a circle centered at (0, 4) with radius \(\sqrt{5}\).

Step 1: Identify the Equation Type

First, we need to identify what type of conic section this equation represents by inspecting the coefficients of the squared terms. The equation is given as \(x^2 - 8y + y^2 + 11 = 0\). Notice that both \(x^2\) and \(y^2\) are present, and both have coefficients of 1 and no cross-terms like \(xy\), suggesting it could be a circle or an ellipse.

Step 2: Rearrange the Equation

Rewrite the equation by grouping the \(x\) and \(y\) terms together to prepare for completing the square. Begin rearranging it to look like this: \(x^2 + y^2 - 8y = -11\).

Step 3: Complete the Square

Complete the square for the \(y\) terms. Start by focusing on the \(y^2 - 8y\) part. To complete the square, take half of the coefficient of \(y\), square it, and then add and subtract that value within the equation: \[y^2 - 8y = (y - 4)^2 - 16\].

Step 4: Rewrite the Equation in Standard Form

Substitute the completed square into the equation: \[x^2 + (y - 4)^2 - 16 = -11\]. Simplify it to bring it into standard form by adding 16 to both sides: \[x^2 + (y - 4)^2 = 5\]. This resembles the equation of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2 = 5\), \(h = 0\), and \(k = 4\).

Step 5: Identify the Conic Section

The equation \(x^2 + (y - 4)^2 = 5\) is in the standard form of a circle, pointing out that it is a circular conic section with center \((0, 4)\) and radius \(\sqrt{5}\).

Step 6: Graph the Equation

Use the information about the circle to draw the graph. The center of the circle is at the point \((0, 4)\), and its radius is \(\sqrt{5} \approx 2.24\). Plot the center and draw a circle around it extending approximately 2.24 units in all directions from the center.

Key Concepts

Circle

A circle is one of the fundamental shapes in geometry. It represents all points that are a fixed distance, known as the radius, from a central point, called the center. In terms of equations, the circle's equation in standard form is given by \[(x - h)^2 + (y - k)^2 = r^2\] Where:

• \( h \) and \( k \) are the x and y coordinates of the circle's center.
• \( r \) is the radius, which will always be a positive number as it represents distance.

In the solution, we transformed the original conic equation into the standard form of a circle's equation. From this new equation, we can clearly identify the center and radius, making it easier to graph the circle.

Completing the Square

Completing the square is a mathematical technique used to transform quadratic expressions into a perfect square trinomial, which simplifies the solving process. It is especially useful when converting a quadratic equation from general form to standard form. Here's how you complete the square:

• Identify the quadratic term and linear term, focusing on one variable at a time.
• Take half of the linear term's coefficient and square it.
• Add and subtract this square inside the equation, thereby creating a perfect square trinomial which can be factored easily.

In our specific exercise, we dealt with the terms \(y^2 - 8y\). By taking half of -8 (which is -4) and squaring it (to get 16), we write it as \[(y - 4)^2 - 16\]. This transformation allows the equation to be rearranged in standard form, revealing more about the conic section.

Standard Form

The standard form is an organized way of expressing the equation of conic sections like parabolas, circles, ellipses, and hyperbolas. It places key information such as the center of the circle, the radius, or the vertices and foci of an ellipse clearly within the equation. For circles, as discussed:

• The standard form is \((x - h)^2 + (y - k)^2 = r^2\).
• This positioning makes it very straightforward to read off the circle's center and radius directly from the equation.

In our exercise, rewriting the equation to \[x^2 + (y - 4)^2 = 5\] connects the abstract numbers in the equation to a clear, visual shape once graphed. Standard form simplifies both interpretation and graphing, which is invaluable in both learning and applying mathematics.