Problem 40 - College Algebra

Question

Sketch a graph of the hyperbola, labeling vertices and foci. \(-x^{2}+8 x+4 y^{2}-40 y+88=0\)

Step-by-Step Solution

Answer

This hyperbola is centered at (4,5) with vertices at approximately (4, 2.35) and (4, 7.65), and foci at (4, -0.92) and (4, 10.92).

Step 1: Rearrange the Equation

First, rearrange the given equation to separate the variables:\[-x^2 + 8x + 4y^2 - 40y + 88 = 0\] becomes\[-(x^2 - 8x) + 4(y^2 - 10y) = -88\].

Step 2: Complete the Square for x and y

Complete the square for both x and y:For the x terms: \[ - (x^2 - 8x) = -(x-4)^2 + 16 \]. For the y terms: \[ 4(y^2 - 10y) = 4((y-5)^2 - 25) = 4(y-5)^2 - 100 \]. Thus, substituting back into the equation, it becomes:\[-(x-4)^2 + 4(y-5)^2 = -88 + 16 + 100\].

Step 3: Simplify the Equation

Simplify the equation:\[-(x-4)^2 + 4(y-5)^2 = 28\]. Divide the entire equation by 28 to make the right side equal to 1:\[\frac{-(x-4)^2}{28} + \frac{4(y-5)^2}{28} = 1\].

Step 4: Rewrite as a Standard Hyperbola Equation

Rewrite the equation in the standard form of a hyperbola:\[-\frac{(x-4)^2}{28} + \frac{(y-5)^2}{7} = 1\]. This can be further simplified to:\[\frac{(y-5)^2}{7} - \frac{(x-4)^2}{28} = 1\]. This is in the form \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\].

Step 5: Identify the Center, Vertices, and Foci

Identify the center, vertices and calculate foci:The center is \((h, k) = (4, 5)\). The vertices are determined from \[a = \sqrt{7} \approx 2.65\], so the vertices are at \((4, 5\pm2.65)\).The foci are determined using \[c^2 = a^2 + b^2 = 7 + 28 = 35\], so \[c = \sqrt{35} \approx 5.92\], which gives foci at \((4, 5\pm5.92)\).

Step 6: Sketch the Hyperbola

Begin sketching the hyperbola by plotting the center at \((4,5)\). Mark the vertices at \((4, 2.35)\) and \((4, 7.65)\). Then mark the foci at \((4, -0.92)\) and \((4, 10.92)\). Draw the branches of the hyperbola opening vertically along the y-axis, centered on the point \((4,5)\).

Key Concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is essential for graphing and analyzing these curves. A hyperbola primarily involves two squared terms, one with a positive coefficient and the other with a negative coefficient, always separated by a minus sign. This form resembles:

• Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Where \( (h,k) \) represents the center of the hyperbola, \( a \) is the distance from the center to the vertices, and \( b \) is the distance that eventually helps determine the foci. Recognizing this standard form allows us to easily extract information such as center coordinates, axis orientation, and the associated geometric properties of the hyperbola.

Completing the Square

Completing the square is a key algebraic technique used to transform a quadratic expression into a perfect square form. This is particularly useful when dealing with conic sections like hyperbolas.

Steps in Completing the Square

To complete the square for an equation:

• Take the quadratic term (e.g., \(x^2 - 8x\)), halve the linear coefficient (-8), which is -4, and square it (16).
• Add and subtract this square inside the equation (\((x-4)^2 - 16\)).
• Repeat this for the \(y\) terms: \(y^2 - 10y\) becomes \((y-5)^2 - 25\).

Once you have these, rearrange them to transform the expression into a standard hyperbola format. This step ensures the equation can be aligned with comparable geometric properties like center and vertices easily.

Centers and Vertices of Hyperbolas

The center and vertices are fundamental components when identifying and sketching a hyperbola. The center acts as the midpoint, around which the hyperbola opens. For the hyperbola in standard form:

Locating the Center

The center \((h, k)\) is calculated directly from the transformed hyperbola equation, here it is \((4, 5)\) derived from \((x-4)^2 \) and \((y-5)^2 \).

Determining the Vertices

Vertices line up on the axis through the center and are located \(a\) units away, vertically for a vertical hyperbola. In this case, the calculation for \(a \) is \(\sqrt{7} \approx 2.65\). Thus, vertices are at \((4, 5 \pm 2.65)\), emphasizing the vertical stretch and span.

Finding Foci of Hyperbolas

The foci of a hyperbola are two points that lie along the axis of a hyperbola, further out than the vertices, and are crucial in defining its shape. The distance from the center to each focus is determined by the relationship \(c^2 = a^2 + b^2\).

Calculating the Foci

For our current hyperbola, with \(a^2 = 7\) and \(b^2 = 28\), sum these to find \(c^2 = 35\), leading to \(c = \sqrt{35} \approx 5.92\).

• This offers foci at \((4, 5 \pm 5.92)\), which are positioned along the y-axis for a vertically oriented hyperbola.

Plotting these alongside the center and vertices helps in constructing an accurate graphical representation of the hyperbola.