Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(x^{2}+2 x-100 y^{2}-1000 y+2401=0\)

Step-by-Step Solution

Answer

The standard form is \(\frac{(x+1)^2}{2400} - \frac{(y+5)^2}{24} = 1\); vertices: \((-50, -5)\) and \((48, -5)\); foci approximately \((-50.04, -5)\) and \((48.04, -5)\); asymptotes are \(y=\pm\frac{1}{10}x - 4.9\).

Step 1: Arrange the terms

Start by arranging the terms of the given equation \(x^{2}+2x-100y^{2}-1000y+2401=0\) so that the \(x\) terms and \(y\) terms are grouped together: \(x^2 + 2x - 100y^2 - 1000y = -2401\).

Step 2: Complete the square for x

To complete the square for the \(x\) terms, take \(x^2 + 2x\): Add and subtract \((\frac{2}{2})^2 = 1\) within the equation. The \(x\) terms become \((x+1)^2 -1\).

Step 3: Complete the square for y

To complete the square for the \(y\) terms, take \(-100(y^2 + 10y)\): Add and subtract \((\frac{10}{2})^2 = 25\) within the equation, keeping the factor of \(100\): The terms become \(-100((y+5)^2 - 25)\). After factoring, it becomes \(-100(y+5)^2 +2500\).

Step 4: Create the standard form of the hyperbola

Substitute the completed square expressions into the equation: \((x+1)^2 - 1 - 100(y+5)^2 + 2500 = -2401\). Simplify to get \((x+1)^2 - 100(y+5)^2 = -2400\). Then, divide the entire equation by \(-2400\): \[\frac{(x+1)^2}{2400} - \frac{(y+5)^2}{24} = 1\] which is in the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).

Step 5: Identify the center, vertices, and foci

The center of the hyperbola is \((-1, -5)\). Since the \(x\)-term has the larger denominator, this hyperbola opens horizontally. Vertices are at \((-1\pm \sqrt{2400}, -5)\) which simplify to approximately \((-50, -5)\) and \((48, -5)\). The distance to the foci \(c\) is determined by \(c^2 = a^2 + b^2 = 2400 + 24 = 2424\), so \(c = \sqrt{2424}\), resulting in foci approximately at \((-1\pm \sqrt{2424}, -5)\).

Step 6: Write equations of asymptotes

The equations of the asymptotes for this hyperbola are based on the slopes \( \pm\frac{b}{a}\), which are derived from \( \pm \frac{\sqrt{24}}{\sqrt{2400}}\). The equations for the asymptotes then are \( y+5 = \pm \frac{1}{10}\cdot(x+1) \), simplifying to \( y = \pm \frac{1}{10}x - 4.9 \).

Key Concepts

Equation of Hyperbola

A hyperbola is a type of conic section that appears as two open, curved branches. Understanding its equation in standard form is pivotal to determine various properties. The standard form of a hyperbola oriented horizontally is:

• \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]

Here, \((h, k)\) represents the center of the hyperbola.
In the exercise, the original equation is transformed into its standard form by completing the square. This technique involves rearranging and grouping \(x\) and \(y\) terms perfectly to factor them easily. Following simplifications, we obtain the equation:

• \[\frac{(x+1)^2}{2400} - \frac{(y+5)^2}{24} = 1\]

This indicates a horizontally oriented hyperbola with the center at \((-1, -5)\).

Vertices of Hyperbola

Vertices are critical points on a hyperbola located on its major axis. They aid in visualizing the hyperbola's orientation and scale. For a hyperbola aligned horizontally like \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\], vertices are computed as \((h \pm a, k)\).
In the given equation, \(a^2 = 2400\), so \(a = \sqrt{2400}\).
The center is at \((-1, -5)\), thus vertices will be at:

• \((-1 \pm \sqrt{2400}, -5)\)
• This calculates approximately to \((-50, -5)\) and \((48, -5)\).

This shows the points where the hyperbola crosses its transverse axis, helping define its spread along the axis. It’s these points that are indispensable to sketching or analyzing the graph of a hyperbola.

Foci of Hyperbola

The foci are another set of defining points located within the hyperbola, which hold unique geometric significance. In a hyperbola, the sum of distances from any point on the curve to the two foci is constant.
For hyperbolas, the distance \(c\) from the center to the foci is calculated using \(c^2 = a^2 + b^2\), where \(a\) and \(b\) are derived from the denominators of the standard form.

• Here, \(a^2 = 2400\) and \(b^2 = 24\) make \(c^2 = 2400 + 24 = 2424\).
• Solving, we find \(c = \sqrt{2424}\), which determines the foci positions.
• Adding and subtracting \(c\) from the \(x\)-coordinate (since it's horizontal) of the center \((-1, -5)\) gives foci approximately at \((-50\pm \sqrt{2424}, -5)\).

Understanding the foci allows for a deeper insight into the geometric properties and behavior of the hyperbola.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches inevitably yet never touches. They are crucial for understanding the general direction and orientation of the hyperbola branches.
For a hyperbola in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the equations for asymptotes are:

• \(y - k = \pm\frac{b}{a}(x - h)\)

In our problem:

• \(a = \sqrt{2400}\) and \(b = \sqrt{24}\).
• Thus, \(\pm\frac{b}{a} = \pm\frac{1}{10}\).

The equations become:

• \(y + 5 = \pm \frac{1}{10}(x + 1)\)
• This simplifies to \(y = \pm \frac{1}{10}x - 4.9\).

Asymptotes help in sketching the hyperbola as they represent the direction in which the branches extend to infinity, lying symmetrically around these lines.