Problem 77 - 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. $$x^{2}-2 x-y^{2}+2 y=4$$

Step-by-Step Solution

Answer

The center is (1, 1) and the vertices are at (3, 1) and (-1, 1).

Step 1: Rearrange the Equation

Start by rearranging the equation. Move the constants to the right side of the equation: \[ x^2 - 2x - y^2 + 2y = 4 \] Add 4 to both sides: \[ x^2 - 2x - y^2 + 2y - 4 = 0 \].

Step 2: Complete the Square

For the terms involving \(x\) and \(y\), complete the square:For \(x\):\[ x^2 - 2x \] needs completing. Add and subtract \((2/2)^2 = 1\) inside the equation:\[ (x^2 - 2x + 1) - 1 \] becomes \((x-1)^2 - 1\).For \(y\):\[ - (y^2 - 2y) \] needs completing. Add and subtract \((2/2)^2 = 1\) inside the equation:\[ -(y-1)^2 + 1 \].

Step 3: Substitute Back

Substitute the squares back into the rearranged equation:\[ (x-1)^2 - 1 - (y-1)^2 + 1 = 4 \].

Step 4: Simplify the Equation

Simplify by canceling the constant terms:\[ (x-1)^2 - (y-1)^2 = 4 \].

Step 5: Write the Standard Form of the Hyperbola Equation

The equation of the hyperbola in standard form is:\[ \frac{(x-1)^2}{4} - \frac{(y-1)^2}{4} = 1 \].

Step 6: Identify the Center and Vertices

The center of the hyperbola is at \((h, k) = (1, 1)\). The distance \(a\) from the center to each vertex in the direction of the \(x\)-axis is 2 (since \(a^2 = 4\)).Thus, the vertices are:\((1 + 2, 1) = (3, 1)\) and \((1 - 2, 1) = (-1, 1)\).

Key Concepts

Standard Form

To understand hyperbolas, it's important to start with the standard form of their equation. Hyperbolas can take different orientations, but the standard form allows us to analyze them consistently. After manipulating and simplifying the original equation, we reached a familiar form:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]In this equation:

• \( (x-h) \) and \( (y-k) \) represent the shifts from the origin along the x-axis and y-axis, respectively.
• \( a^2 \) and \( b^2 \) are constants that dictate the shape and size of the hyperbola.

Rewriting equations to this form helps us quickly identify essential characteristics of the hyperbola, such as the center and vertices.

Center of Hyperbola

The center of a hyperbola is straightforward to locate once you have the equation in standard form. Just like in our problem, the rearranged equation showed the standard form:\[ \frac{(x-1)^2}{4} - \frac{(y-1)^2}{4} = 1 \]From here, identifying the center becomes easy. The center \(h, k\) aligns with the values that zero out the squared terms. Therefore:

• The center is at \((h, k) = (1, 1)\).

Understanding the center is crucial because all other features of the hyperbola, like its vertices, are measured relative to the center.

Vertices of Hyperbola

For hyperbolas, vertices are significant points that show where the hyperbola branches start to curve the most. Once the equation is in standard form, finding vertices becomes manageable. A hyperbola's vertices lie along the axis that has a positive term in the equation.In our expression \( \frac{(x-1)^2}{4} - \frac{(y-1)^2}{4} = 1 \), the vertices are aligned horizontally since the x-term is positive:

• The distance \(a\) from the center to a vertex along this axis can be calculated by taking the square root of \(a^2\). Here, \(a^2 = 4\), so \(a = 2\).
• The vertices are thus \((1+2, 1) = (3, 1)\) and \((1-2, 1) = (-1, 1)\).

Vertices reflect a hyperbola's maximum extent in each primary direction. This concept helps predict the hyperbola's appearance graphically.