Problem 16 - Algebra 2

Question

Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center and the radius. For an ellipse or a hyperbola, give the center and the foci. Sketch the graph. \(y^{2}-x^{2}+6 x-4 y=6\)

Step-by-Step Solution

Answer

The given equation represents a hyperbola with center at point (3, 2) and foci at points (3, 0) and (3, 4).

Step 1: Group terms with common variables together

Group the x's and y's together like this: \(y^2 - 4y = x^2 - 6x + 6\)

Step 2: Complete the square for both x and y terms

To complete the square, take -b/2a (where ax²+bx is the equation form) for both x and y and observe: for y this equals 2 and for x this equals 3, hence the equation becomes \[(y - 2)^2 = -(x - 3)^2 + 6\].

Step 3: Identify the standard form of the conic section

Rewrite equation as \[(y - 2)^2 = -(x - 3)^2\] by subtracting 6 from both sides. This equation matches the standard form of a hyperbola in the y-direction.

Step 4: Identify the center and foci of the hyperbola

The center of the hyperbola is obtained from the equation by taking (h, k) as the center, here (3, 2). For a hyperbola, the distance to the foci c is given by \(c = \sqrt{a^2 + b^2}\) where a is 2 and b is 0, hence c = 2. The foci are then (3, 2±2) = (3, 0) and (3, 4).

Step 5: Sketch the graph of the hyperbola

The hyperbola opens upward and downward. Draw an xy-plane where the branches of the hyperbola pass through the foci points: (3, 0) and (3, 4). The center of the hyperbola is at the point (3, 2). Draw a vertical line passing through the center, indicating the direction of the hyperbola.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that appears when cutting a double cone with a plane in such a way that two separate curves are formed. A hyperbola is defined by two curves called branches, which are mirror images of each other. The general equation of a hyperbola is either \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) or \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center of the hyperbola. Hyperbolas have characteristic elements such as asymptotes, which are lines that the hyperbola approaches but never touches, and foci, which are points used to define the shape. Each branch of the hyperbola has a corresponding focus, and the distance between the center and a focus is called the focal distance. This focal distance is calculated using the formula \(c = \sqrt{a^2 + b^2}\). In the specific exercise, the equation of the hyperbola is \((y - 2)^2 = -(x - 3)^2\), identifying its center as \((3, 2)\) and indicating that it opens upwards and downwards.

Completing the Square

Completing the square is a method used in algebra to transform quadratic equations into a form that is easier to solve or interpret. This process is particularly useful in conic sections for converting equations into their standard forms. To complete the square, take the quadratic term and constant term, and adjust them to form a perfect square trinomial.
Here’s a simple approach:

• Identify the quadratic expression that needs to be completed.
• For an expression like \(x^2 + bx\), add and subtract \((\frac{b}{2})^2\) such that \((x - \frac{b}{2})^2 = x^2 + bx + (\frac{b}{2})^2\).
• Apply the same method to the y-term as needed. This makes identifying vertices and centers straightforward.

For instance, in the original equation \(y^2 - 4y = x^2 - 6x + 6\), completing the square helps transform it into \((y - 2)^2 = -(x - 3)^2 + 6\). By doing so, understanding the hyperbola's position and orientation becomes clearer, leading to simpler sketching and analysis.

Graphing Conic Sections

Graphing conic sections involves plotting the curves represented by certain types of equations. These sections include parabolas, circles, ellipses, and hyperbolas. Each conic section has its unique equation form and characteristics.
When graphing these sections, it's important to:

• Identify the type of conic from its equation form.
• Extract key features like centers, vertices, axes lengths, and directions of opening.
• Determine parameters like foci or directrices, which further guide the drawing.

In the case of a hyperbola, such as the one derived from the exercise, the graph involves plotting its center and using its standard form to determine the orientation and direction of the branches. Focusing on the hyperbola given by \((y - 2)^2 = -(x - 3)^2\), we recognize the center at \((3, 2)\) and draw the branches oriented vertically, acknowledging the foci positions at \((3, 0)\) and \((3, 4)\). Sketching this graph accurately helps visualize the hyperbola's positive and negative ranges, demonstrating the hyperbola's symmetry and asymptotic behavior.