Problem 23 - College Algebra

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

Equation: \( \frac{(x + 1)^2}{2525.01} - \frac{(y + 5)^2}{2524.01} = 1 \), vertices: \((-1, -5 \pm 1)\), foci: \((-1, -5\pm \sqrt{2525.01})\).

Step 1: Rearrange and Group Terms

Start by grouping the terms involving \(x\) together and the terms involving \(y\) together. This gives you:\[x^2 + 2x - 100(y^2 + 10y) + 2401 = 0\]

Step 2: Complete the Square

To complete the square for \(x\), take half of the coefficient of \(x\), square it, and add it inside the group. For \(x\), this gives \((2/2)^2 = 1\). Add and subtract 1:\[(x^2 + 2x + 1 - 1) - 100(y^2 + 10y) + 2401 = 0\]Now for \(y\), take half of 10 which is 5, square it to get 25, add and subtract 2500 in the group (remember to multiply by 100 since the equation is \(-100(y^2 + 10y)\)):\[(x^2 + 2x + 1) - 100(y^2 + 10y + 25 - 2500) + 2401 = 0\]

Step 3: Simplify the Equation

Rewrite the completed squares:\[(x + 1)^2 - 100((y + 5)^2 - 2500) + 2401 = 0\]Distribute and simplify:\[(x + 1)^2 - 100(y + 5)^2 + 250000 + 2401 = 0\]Combine constants:\[(x + 1)^2 - 100(y + 5)^2 + 250000 + 2401 = 0\]\[(x + 1)^2 - 100(y + 5)^2 + 252401 = 0\]

Step 4: Move Constants

Subtract 252401 from both sides:\[(x + 1)^2 - 100(y + 5)^2 = -252401\]

Step 5: Divide by Constant

Divide the equation by 252401 to get ones on the right side:\[\frac{(x + 1)^2}{252401} - \frac{(y + 5)^2}{2524.01} = -1\]This equation shows a hyperbola in standard form:

Step 6: Identify Vertices and Foci

The center of the hyperbola is \((-1, -5)\). Since \(x\) terms are positive, it's vertical.- Vertices are at \((-1, -5 \pm 1)\).- \(c=\sqrt{1+2524.01}=\sqrt{2525.01}\) which gives foci at \((-1, -5\pm \sqrt{2525.01})\).

Step 7: Asymptotes Calculation

The asymptotes have the form \(y = k \pm m(x-h)\). For ”a” and ”b”: \(a^2 = 1,\) \(b^2 = 2524.01\), \(m = \frac{b}{a}=\sqrt{2524.01}\).The equations are:\[y + 5=\pm \sqrt{2524.01}(x + 1)\]

Step 8: Final Equation Form

After reviewing, the correct interpretation and simplification of initial result leads to: \[\frac{(x + 1)^2}{2525.01} - \frac{(y + 5)^2}{2524.01} = 1\] in standard form, correcting signage found while evaluating intermediate results.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola play a crucial role in revealing its basic shape and orientation. In any hyperbola, the vertices are the points where the hyperbola crosses its transverse axis. These points are located at a specific distance from the center of the hyperbola. To identify the vertices, you need to understand the equation in its standard form.

For the hyperbola equation \[\frac{(x + 1)^2}{2525.01} - \frac{(y + 5)^2}{2524.01} = 1\],
the center is \((-1, -5)\). The constants under the squared terms, in this case, dictate the distance to the vertices from the center.

• The term under \((x + 1)^2\) is \(2525.01\) which is related to the \(x\) axis.
• The vertices along the transverse axis (if hyperbola is horizontal) are calculated by the formula: \((h \pm a, k)\), where \(a\) is the square root of the numerator under the \(x\) term.
• In this hyperbola, as it is vertical due to the larger term under \(y\), the vertices are at \((-1, -5 \pm 1)\).

This concept allows you to visualize the orientation and extent of the hyperbola in its graph.

Foci of Hyperbola

The foci of a hyperbola offer insights into its eccentricity, helping us understand its shape better. Located along the transverse axis, the foci lie at a distance greater than the vertices from the center.

Understanding the role of the foci involves some calculation:

• To find the distance to the foci, use the formula \(c = \sqrt{a^2 + b^2}\). This is because the hyperbola's dimensions dictate \(c\), the distance from the center to each focus, by the relationship between the distance to its vertices and its foci.
• From the equation \[\frac{(x + 1)^2}{2525.01} - \frac{(y + 5)^2}{2524.01} = 1\], we determine \(a^2 = 1\) and \(b^2 = 2524.01\).
• Thus, \(c = \sqrt{1 + 2524.01} = \sqrt{2525.01}\).

This means the foci are located at the coordinates \((-1, -5 \pm \sqrt{2525.01})\), revealing points beyond the vertices on each side along the transverse axis, further defining the hyperbola's stretched appearance.

Asymptotes of Hyperbola

The asymptotes of a hyperbola are diagonal lines that the hyperbola approaches but never intersects. These lines help define the hyperbola's open and infinite nature by providing a visual boundary that the curve will never cross.

Finding the equations of the asymptotes involves some steps:

• For the hyperbola equation \[\frac{(x + 1)^2}{2525.01} - \frac{(y + 5)^2}{2524.01} = 1\], notice \(a = \sqrt{1}\) and \(b = \sqrt{2524.01} = b\).
• The slopes of the asymptotes are given by \(\pm \frac{b}{a}\) for a hyperbola aligned with the axes.
• In this example, \(b = \sqrt{2524.01}\) and \(a = 1\), giving a slope \(m = \pm \sqrt{2524.01}\).
• The asymptotes' equations, based on the slope and the center position, are: \(y + 5 = \pm \sqrt{2524.01}(x + 1)\).

These asymptotes provide a geometric boundary for the hyperbola, allowing us to better predict and understand the behavior of its arms as they infinitely stretch.