Problem 79 - A Graphical Approach to Precalculus with Limits

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$3 y^{2}+24 y-2 x^{2}+12 x+24=0$$

Step-by-Step Solution

Answer

The center is (3, -4); vertices are (3, -2) and (3, -6).

Step 1: Rearrange and Group Terms

Rearrange the given equation to group the y-terms together and the x-terms together: \[ 3y^{2} + 24y - 2x^{2} + 12x + 24 = 0 \] becomes \[ 3(y^{2} + 8y) - 2(x^{2} - 6x) = -24 \].

Step 2: Complete the Square for Y-terms

To complete the square for the y-terms, take \( y^{2} + 8y \), add and subtract \((\frac{8}{2})^2 = 16\): \[ y^{2} + 8y = (y+4)^2 - 16 \]. Substitute back into the equation to get: \[ 3((y+4)^2 - 16) \].

Step 3: Complete the Square for X-terms

For the x-terms, take \( x^{2} - 6x \), add and subtract \((\frac{6}{2})^2 = 9\): \[ x^{2} - 6x = (x-3)^2 - 9 \]. Substitute back into the equation to get: \[ -2((x-3)^2 - 9) \].

Step 4: Simplify and Set Equation to Zero

Expand and simplify both completed square expressions: \[ 3((y+4)^2 - 16) - 2((x-3)^2 - 9) = -24 \]. This becomes \[ 3(y+4)^2 - 48 - 2(x-3)^2 + 18 = -24 \].

Step 5: Solve for Hyperbola Equation

Combine like constants and rearrange the equation: \[ 3(y+4)^2 - 2(x-3)^2 = 6 \]. Divide the entire equation by 6 to obtain the standard form: \[ \frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1 \].

Step 6: Identify Center and Vertices

The equation \( \frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1 \) identifies the hyperbola centered at \((h, k) = (3, -4)\). The vertices are \((3, -2)\) and \((3, -6)\), found by solving \(y+4=\pm\sqrt{2}\).

Key Concepts

Standard Form

The standard form of a hyperbola's equation provides a clear structure, making it easier to understand and work with.
In general, the standard form for a hyperbola can be one of the following, depending on its orientation:

• Horizontal hyperbola: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]
• Vertical hyperbola: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]

Here, \( (h, k) \) is the center of the hyperbola, and \( a \) and \( b \) are real numbers that determine the shape and placement of the hyperbola's vertices.
This allows one to quickly identify special points and graph the hyperbola more easily. For the given problem, the example results in the standard form \[\frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1\].
This showcases a vertical hyperbola centered at \((3, -4)\).

Completing the Square

Completing the square is a technique used to transform a quadratic equation into its graph-friendly form. This method allows us to rewrite a quadratic polynomial into a perfect square trinomial.
In the given exercise, this process was applied separately to both the \( y \) and \( x \) terms:

• For the \( y \) terms \((y^2 + 8y)\), adding and subtracting \(16\) allowed us to express it as \((y+4)^2 - 16\).
• Similarly, for the \( x \) terms \((x^2 - 6x)\), adding and subtracting \(9\) resulted in \((x-3)^2 - 9\).

After substituting these completed squares back into the equation, we had convenient quadratic expressions to work with. This step is crucial as it transforms the equation into one that can be easily manipulated into the standard form of a hyperbola.

Center and Vertices

For hyperbolas, identifying the center and vertices is essential as they define the shape and position of the graph.
The center \((h, k)\) acts as the midpoint of the hyperbola, and this problem revealed it to be at \((3, -4)\).
Vertices are the furthest points on the hyperbola from the center on the axis. These are especially important as they determine the opening of the hyperbola.

• In this case, the vertices are found vertically aligned above and below the center – calculated by adjusting the \(y\) component.
• Specifically, the vertices include the points \((3, -2)\) and \((3, -6)\), derived by finding values that satisfy the equation \(y+4=\pm\sqrt{2}\) in the final standard form equation.

Understanding these points is key to sketching a hyperbola accurately as they provide both direction and proportion of the curve.