Problem 27 - Algebra 2

Question

Sketch the graph of each equation. $$ x^{2}-4 y^{2}+2 x+24 y=51 $$

Step-by-Step Solution

Answer

The gamma of the equation \(x^{2}-4 y^{2}+2 x+24 y=51\) is a hyperbola with center at (-1,3), vertices at (-1±√52, 3), co-vertices at (-1, 3±√13) and equations of the asymptotes are \( y - 3 = \pm (\frac{\sqrt{13}}{\sqrt{52}})(x + 1) \).

Step 1: Rewrite the given equation

The given equation \(x^{2}-4 y^{2}+2 x+24 y=51\) must be rewritten. Start by grouping the x terms and y terms respectively and completing the square. The equation thus becomes: \( (x^2 + 2x + 1) - 4(y^2 - 6y + 9) = 51 + 1 - 4 \times 9 \). This simplifies to \( (x+1)^2 - 4(y-3)^2 = 52 \)

Step 2: Convert into standard form

To convert the equation into the standard form for a hyperbola, it's needed to divide by 52 on both sides \( \frac{(x+1)^2}{52} - \frac{(y-3)^2}{13} = 1 \)

Step 3: Identify the center, vertices, co-vertices and asymptotes

In the standard form, (h, k) gives the center of the hyperbola. Here, the center is (-1, 3). The value under x, i.e., 52, is \(a^2\), so \(a = \sqrt{52}\), and similarly, the value under y, i.e., 13, is \(b^2\), so \(b = \sqrt{13}\). So, the vertices would be (-1±√52, 3) and co-vertices would be (-1, 3±√13). The equations of asymptotes are \( y - k = \pm (\frac{b}{a})(x - h) \), which translate to \( y - 3 = \pm (\frac{\sqrt{13}}{\sqrt{52}})(x + 1) \)

Step 4: Sketch the hyperbola

First, plot the center point, then draw the guiding rectangle using the distances a and b. The rectangle is centered at the hyperbola's center. Plot the asymptotes which pass through the corners of the rectangle, and finally sketch the hyperbola that approaches but never touches these asymptotes. The two arms of the hyperbola would lie inside the rectangle and cross through the vertices.

Key Concepts

Graphing conic sections

Understanding how to graph conic sections, like hyperbolas, is crucial in many areas of mathematics. Conic sections include circles, ellipses, parabolas, and hyperbolas, each with unique properties and equations. To graph a specific conic section, you first need to manipulate its equation into a recognizable form - often the standard form for that conic.
This process might involve rewriting and rearranging the equation, as well as possibly completing the square. For hyperbolas, the graph typically consists of two disconnected curves or "branches", which open either horizontally or vertically depending on the terms of the equation.
The center, vertices, and asymptotes of a hyperbola are essential to plotting its graph. By marking these features on the coordinate plane, you set the stage to sketch the hyperbola accurately, ensuring the branches approach but never actually touch the asymptotes.

Completing the square

Completing the square is a method used in algebra to transform a quadratic equation into a perfect square trinomial. This can make graphing and understanding conic sections easier.
In the problem at hand, we complete the square for both x and y terms separately. Initially, you group the x terms and y terms from the equation: \( x^2 + 2x \) and \( -4y^2 + 24y \).
After completing the square, the equation becomes \((x+1)^2 - 4(y-3)^2 = 52\). Here, completing the square helped convert the original equation into a form that reveals the nature of the conic section more naturally, preparing it for the next step where we put it into the standard form for a hyperbola.

Hyperbola standard form

The standard form of a hyperbola's equation helps us identify important characteristics, such as its center, vertices, and asymptotes. The equation for a hyperbola generally looks like \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) for horizontal hyperbolas, or \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) for vertical ones.
Here, \((h, k)\) represents the center of the hyperbola.
In our example, we transformed the equation to \( \frac{(x+1)^2}{52} - \frac{(y-3)^2}{13} = 1 \), identifying the center as \((-1, 3)\). From here, the values of \(a^2\) and \(b^2\) help determine the distances of vertices and asymptotes from the center, pivotal in sketching the hyperbola.

Asymptotes of a hyperbola

Asymptotes play a crucial role in the shape of a hyperbola. They are lines that the hyperbola approaches but never actually intersects. These asymptotes outline an "X"-shape that guides the direction of the hyperbola's branches.
To find the equations of the asymptotes, use the slopes derived from \( \pm \frac{b}{a} \), where \( a \) and \( b \) are the distances associated with the standard form of the hyperbola.
In our equation, the asymptotes are \( y - 3 = \pm \left(\frac{\sqrt{13}}{\sqrt{52}}\right)(x + 1) \), which provide the lines along which the branches of the hyperbola curve. Graphing the asymptotes first gives a visual structure that helps in sketching the hyperbola accurately, ensuring that the branches correctly approach these guiding lines.