Problem 45 - Algebra and Trigonometry

Question

convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$ 16 x^{2}-y^{2}+64 x-2 y+67=0 $$

Step-by-Step Solution

Answer

The standard equation of the hyperbola is \((x + 2)^{2}/(-48) - (y + 1)^{2}/48 = 1\). The center is (-2, -1), the vertices are (-2-√48, -1) and (-2+√48, -1), the foci are √2*48 units away from the center in the horizontal direction, and the equations of the asymptotes are \(y = -1 \pm (x + 2)\)

Step 1: Grouping Like Terms

Observe the equation and group the x-terms and the y-terms together:\(16x^{2} + 64x - y^{2} - 2y + 67 = 0\)Re-order the equation to get:\(16x^{2} + 64x - (y^{2} + 2y) = -67\)

Step 2: Completing the Square

We now complete the square for both x and y. For the x-term group, divide by 16 first to simplify:\((x^{2} + 4x) - (y^{2} + 2y) = -67/16\)Add square of half of the coefficient of x, in this case 4/2 = 2:\((x^{2} + 4x + 4) - (y^{2} + 2y) = -67/16 + 4\)Now, for the y-term group, add the square of half the coefficient of y, in this case 2/2 = 1:\((x^{2} + 4x + 4) - (y^{2} + 2y + 1) = -67/16 + 4 - 1\)Simplify the equation to:\((x + 2)^{2} - (y + 1)^{2} = -3/16\)

Step 3: Normalize the Hyperbola Equation

A standard form of a hyperbola is \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), so normalize the equation:\((x + 2)^{2}/(-3/16) - (y + 1)^{2}/(-3/16) = 1\)Solve it to:\((x + 2)^{2}/(-48) - (y + 1)^{2}/(-48) = 1\)This further simplifies to:\((x + 2)^{2}/(-48) - (y + 1)^{2}/48 = 1\)

Step 4: Find the Parameters of the Hyperbola

From the equation, we can identify the center of the hyperbola as (-2, -1). The semi-major axis (a) is √48 and the semi-minor axis (b) is also √48. The foci can be found using the equation \(c = √(a² + b²)\), which will yield c = √2*48. The vertices are located at a distance of 'a' from the center along the x-axis, so vertices are (-2-√48, -1) and (-2+√48, -1).

Step 5: Find the Asymptotes of the Hyperbola

The equations of the asymptotes in standard form are given by \(y = k \pm (b/a)(x - h)\). For this hyperbola, \(y = -1 \pm (x + 2)\).

Key Concepts

Completing the Square

Completing the square is a method used to rewrite quadratic expressions in a form that can be easily analyzed and solved. Here, the goal is to transform the equation of a hyperbola into its standard form.

For our equation, we start with grouping the x-terms and the y-terms separately. We then adjust these grouped terms by adding and subtracting specific values to form perfect squares. This helps simplify the equation into a recognizable form.

**Steps to Complete the Square:**

• Take the coefficient of the x-term, divide it by 2, and square it.
• Add and subtract this squared value inside the equation.
• Do the same for the y-terms.

The resulting equation forms the basis for further simplification into the standard form of a hyperbola.

Standard Form of Hyperbola

The standard form of a hyperbola is essential as it reveals key characteristics of the hyperbola, including its center, vertices, and orientation. The basic form is given by: \[ (x-h)^2/a^2 - (y-k)^2/b^2 = 1 \]

Where \(h, k\) are the coordinates of the center, and \(a\) and \(b\) are the distances on the major and minor axes respectively.

**Understanding the Components:**

• **Center:** The point \(h, k\).
• **Vertices:** Located along the transverse axis, at a distance of \(a\) units from the center.
• **Orientation:** If \(a^2\) is under the x-term, the hyperbola opens left-right; if under the y-term, it opens up-down.

This form makes graphing and identifying other properties simpler.

Foci of Hyperbola

The foci of a hyperbola are two points that lie on its major axis, one inside each "branch" of the hyperbola. Their key characteristic is that the difference in distances from any point on the hyperbola to these two points is constant.

**Finding the Foci:**

• Use the formula: \(c^2 = a^2 + b^2\)
• Calculate \(c\) to find the position of the foci along the transverse axis.

For example, in our exercise:

If \(a^2 = 48\), then \(b^2 = 48\), and thus \(c = \sqrt{96}\), giving us the distance from the center to each focus.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches. They provide a framework that defines the "opening" of the hyperbola.

In the standard form, the equations of the asymptotes can be derived as:
\( y = k \pm \frac{b}{a}(x - h) \)

**Key Components:**

• **Slope:** The slope \(\pm \frac{b}{a}\) indicates the steepness and direction of the asymptotes.
• **Intersect:** They pass through the center \(h, k\).

For the given exercise, the asymptotes are \(y = -1 \pm (x + 2)\), illustrating how they guide the shape of the hyperbola.