Problem 29 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ 4 x^{2}-25 y^{2}-8 x-96=0 $$

Step-by-Step Solution

Answer

Vertices: (6, 0) and (-4, 0). Foci: (1±√29, 0). Asymptotes: y=±2/5(x-1).

Step 1: Rewrite the Equation

Start by rewriting the given equation in standard form. First, group the x and y terms together: \\[ 4x^2 - 8x - 25y^2 = 96. \]

Step 2: Complete the Square for x

To complete the square for the x terms, factor out a 4 from \( x^2 - 2x \) to get: \ \[ 4(x^2 - 2x) - 25y^2 = 96. \] \ Now complete the square inside the parenthesis: \ \[ x^2 - 2x = (x - 1)^2 - 1. \] \ Substitute this back into the equation: \ \[ 4((x - 1)^2 - 1) - 25y^2 = 96. \] \ This simplifies to: \ \[ 4(x - 1)^2 - 4 - 25y^2 = 96. \]

Step 3: Simplify the Equation

Add 4 to both sides to balance the equation: \ \[ 4(x - 1)^2 - 25y^2 = 100. \] \ Divide the entire equation by 100 to set it equal to 1: \ \[ \frac{(x - 1)^2}{25} - \frac{y^2}{4} = 1. \]

Step 4: Identify the Center, Vertices, and Foci

This equation is now in the standard form \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \) where \( a^2 = 25 \) and \( b^2 = 4 \). \ The center of the hyperbola is \((h, k) = (1, 0)\). \ The vertices are given by \((h \pm a, k)\), so \(a = 5\), leading to vertices at \((1 \pm 5, 0)\) or \((6, 0)\) and \((-4, 0)\). \ The foci are given by \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2} = \sqrt{25 + 4} = \sqrt{29}\), leading to foci at \((1 \pm \sqrt{29}, 0)\).

Step 5: Determine Equations of the Asymptotes

The asymptotes for the hyperbola in this form are given by the equations: \ \[ y = \pm \frac{b}{a} (x-h) + k. \] \ Since \(b = 2\) and \(a = 5\), this gives: \ \[ y = \pm \frac{2}{5} (x-1). \] \ So the equations of the asymptotes are \( y = \frac{2}{5}(x-1) \) and \( y = -\frac{2}{5}(x-1) \).

Key Concepts

Vertices

Vertices are crucial points on a hyperbola as they represent the points where the hyperbola is closest to or intersects the center line of symmetry of the graph. These are the 'turning points' of the curve. For a hyperbola in standard form like the one given in the solution, the formula to find the vertices is

• For a horizontally oriented hyperbola such as \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \), the vertices are located at \((h \pm a, k)\).
• In our example, since the center \((h, k)\) is \((1, 0)\) and \(a = 5\), the vertices can be found at \((1 \pm 5, 0)\), resulting in the points \((6, 0)\) and \((-4, 0)\).

The vertices should be plotted as they essentially outline the width of the hyperbola. Being able to locate these accurately helps in sketching the initial framework of the hyperbola on a graph.

Foci

The foci (singular: focus) are points inside the hyperbola that possess defining properties for the shape. In relation to vertices, the foci lie along the transverse axis and are crucial in forming the hyperbola. The distance from the center to each focus is denoted as \(c\). Here’s how they function:

• The formula to determine the distance to the foci (\(c\)) from the center is \(c = \sqrt{a^2 + b^2}\).
• In this solution, we calculate \(c = \sqrt{25 + 4} = \sqrt{29}\), which places the foci at \((1 \pm \sqrt{29}, 0)\).

Locating the foci is important as it helps describe the set of points that form the hyperbola — the difference between the distances from any point on the hyperbola to each focus is constant, a fundamental property that helps distinguish hyperbolas from other conic sections.

Asymptotes

Asymptotes for hyperbolas are lines that the curve approaches but never reaches, providing a guide for the hyperbola's shape as it stretches to infinity. They form a cross through the center, giving a visual framework of the hyperbola:

• For hyperbolas of the form \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \), the asymptotes are given by the equations \( y = \pm \frac{b}{a} (x-h) + k \).
• In our exercise, with \(b = 2\) and \(a = 5\), the asymptotes are described by the equations \( y = \frac{2}{5}(x-1) \) and \( y = -\frac{2}{5}(x-1) \).

Drawing these lines on a graph before attempting to sketch the hyperbola itself is a smart approach. It allows you to visualize how the hyperbola approaches these lines as it extends outwards, helping to ensure the accuracy of the graph's shape.