Identifying the Conic Section: A Step-by-Step Guide

This article will guide you through the process of identifying the type of conic section represented by the equation (5x^2 - 5y^2 2 - 40y - 10 0). We'll break down the steps to transform this equation into a more recognizable form and identify its type. By the end, you'll be able to determine whether it represents a circle, ellipse, hyperbola, or another conic section.

Transforming the Equation

Let's start by examining the given equation:

(5x^2 - 5y^2 2 - 40y - 10 0)

Step 1: Simplify and Rearrange

First, simplify the equation by combining constants:

(5x^2 - 5y^2 - 40y - 8 0)

Next, move the constant term to the right-hand side (RHS) to isolate the variables:

(5x^2 - 5y^2 - 40y 8)

Now, to understand the type of conic section, observe that the equation lacks an (xy) term. This means the axes of the curve are parallel to the coordinate axes, which simplifies the process of identifying its type.

Step 2: Completing the Square

To complete the square for the (y) terms, we first need to divide the entire equation by 5 to simplify it further:

(x^2 - y^2 - 8y frac{8}{5})

Next, complete the square for the (y) terms:

(x^2 - (y^2 8y) frac{8}{5})

Add and subtract the square of half the coefficient of (y) inside the parentheses:

(x^2 - (y^2 8y 16 - 16) frac{8}{5})

Simplify to get:

(x^2 - (y 4)^2 16 frac{8}{5})

Moving the constant to the RHS:

(x^2 - (y 4)^2 frac{8}{5} - 16)

Simplify the right-hand side:

(x^2 - (y 4)^2 frac{8 - 80}{5})

(x^2 - (y 4)^2 -frac{72}{5})

Step 3: Identifying the Type of Conic Section

The standard form of a conic section is (ax^2 by^2 cx dy e 0). In our transformed equation, we observe:

(x^2 - (y 4)^2 -frac{72}{5})

The coefficients of (x^2) and ((y 4)^2) are (1) and (-1) respectively, indicating that (ab

However, it's important to consider the special cases:

If (a b), then it's a circle. If (a eq b), then it's a hyperbola. If (a b 0), then it's a parabola.

Since (a 1) and (b -1), we identify that it is a hyperbola.

Special Case: Circle

Ignoring the linear terms, the equation simplifies to (5x^2 - 5y^2 8). Dividing through by 5:

(x^2 - y^2 frac{8}{5})

This represents a circle centered at the origin and scaled by a factor of (sqrt{frac{8}{5}}).

Complete Square to Find the Center and Radius

Completing the square for the simplified equation to find the center and radius:

(x^2 - (y^2 8y 16 - 16) frac{8}{5})

(x^2 - (y 4)^2 16 frac{8}{5})

(x^2 - (y 4)^2 frac{8}{5} - 16)

(x^2 - (y 4)^2 -frac{72}{5})

Moving constants to one side:

(x^2 - (y 4)^2 -frac{72}{5})

The center of the circle is ((-2, -4)) and the radius is (sqrt{22}).

Conclusion

In summary, the given equation (5x^2 - 5y^2 2 - 40y - 10 0) is a hyperbola. If we were to simplify it further, we would find that it represents a circle with a specific center and radius.