Problem 15 - Precalculus: Functions and Graphs

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$4 y^{2}-x^{2}+40 y-4 x+60=0$$

Step-by-Step Solution

Answer

Vertices: \((-2, 5 \pm \sqrt{11})\). Foci: \((-2, 5 \pm \sqrt{55})\). Asymptotes: \(y = 5 \pm \frac{1}{2}(x + 2)\).

Step 1: Rewrite the Hyperbola Equation

First, rearrange the terms in the given equation to group the x terms and the y terms together: \[4y^2 - 40y - x^2 - 4x + 60 = 0\] Move the constant term to the other side: \[4y^2 - 40y - x^2 - 4x = -60\]

Step 2: Complete the Square for x and y

For the x terms: \[-x^2 - 4x\] can be rewritten as \[-((x+2)^2 - 4)\].For the y terms: \[4(y^2 - 10y)\].Complete the square for y: \[y^2 - 10y = (y - 5)^2 - 25\].Hence, \[4((y-5)^2 - 25) = 4(y-5)^2 - 100\].

Step 3: Simplify and Form the Standard Hyperbola Equation

Substitute the completed squares back into the equation: \[-((x+2)^2 - 4) + 4(y-5)^2 - 100 = -60\]which simplifies to:\[-(x+2)^2 + 4y^2 - 100 + 4 = -60\]that further simplifies to: \[-(x+2)^2 + 4(y-5)^2 = 44\]Divide every term by 44 to get the standard form: \[\frac{(y-5)^2}{11} - \frac{(x+2)^2}{44} = 1\], which is a hyperbola opening along the y-axis.

Step 4: Identify the Vertices

The standard form corresponds to:\[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\], where the center is at \((h, k) = (-2, 5)\), and \(a^2 = 11\), thus \(a = \sqrt{11}\).The vertices are \((h, k \pm a) = (-2, 5 \pm \sqrt{11})\).

Step 5: Find the Foci

Use the relationship \(c^2 = a^2 + b^2\) to find \(c\), where \(b^2 = 44\). We get:\[c^2 = 11 + 44 = 55\]\(c = \sqrt{55}\).The foci are \((h, k \pm c) = (-2, 5 \pm \sqrt{55})\).

Step 6: Determine the Asymptotes

The equations of the asymptotes for a hyperbola of the form \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]are \[y = k \pm \frac{a}{b}(x-h)\].This gives us:\[y - 5 = \pm \frac{\sqrt{11}}{\sqrt{44}}(x + 2)\], which simplifies to:\[y = 5 \pm \frac{1}{2}(x + 2)\].

Step 7: Sketch the Hyperbola

Draw the center at \((-2, 5)\) and plot the vertices at \((-2, 5 \pm \sqrt{11})\).Plot the foci at \((-2, 5 \pm \sqrt{55})\) and draw the asymptotes as lines that pass through the center with slopes of \(\frac{1}{2}\) and \(-\frac{1}{2}\).Sketch the hyperbola, making sure it approaches the asymptotes.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are crucial points that determine the shape and orientation of the curve. In a standard hyperbola equation like \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center is located at \((h, k)\). The term \(a^2\) corresponds to the distance from the center to each vertex along the transverse axis.
The vertices are found at \((h, k \pm a)\) or \((h \pm a, k)\) depending on whether the hyperbola opens vertically or horizontally.

• In this exercise, the center is at \((-2, 5)\) and \(a = \sqrt{11}\).
• This tells us the hyperbola opens vertically, so the vertices are \((-2, 5 \pm \sqrt{11})\).

Understanding the location of the vertices helps in sketching the hyperbola and understanding its direction of opening.

Foci of Hyperbola

Foci (plural of focus) are integral for understanding the shape of a hyperbola. They are located further from the center than the vertices. The distance from the center to the foci is represented by \(c\), where \(c^2 = a^2 + b^2\).
These points are not on the actual hyperbola but lie along the same axis as the vertices. The role of the foci is in defining the geometry of the hyperbola as a locus of points where the difference in distances to each focus is constant.

• In this specific hyperbola, with \(a^2 = 11\) and \(b^2 = 44\), we calculate \(c^2 = 55\).
• Thus, \(c = \sqrt{55}\), and the foci are located at \((-2, 5 \pm \sqrt{55})\).

Locating the foci is important both for sketching and for understanding the hyperbola's geometric properties.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never actually meets. They are vital for understanding the behavior of the hyperbola at infinity.
For a hyperbola in the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the equations of the asymptotes are derived as \( y = k \pm \frac{a}{b}(x-h) \). These lines provide guidelines for sketching the hyperbola's opening.

• Here, the center \((-2, 5)\), gives the asymptotes \(y-5 = \pm \frac{\sqrt{11}}{\sqrt{44}}(x+2)\).
• Simplifying this, we have \( y = 5 \pm \frac{1}{2}(x+2) \).

These equations help visualize how the hyperbola approaches these lines as it extends outward.

Complete the Square

Completing the square is a technique used to convert quadratic expressions into a perfect square trinomial, facilitating the transition to the standard form of a conic section. For hyperbolas, this method helps arrange terms for easy identification of the equation's parameters.

• In this exercise, we complete the square for both \(x\) and \(y\) terms to rewrite the given hyperbola equation.
• For \(x\): \(-x^2 - 4x\) becomes \(-( (x+2)^2 - 4)\).
• For \(y\): \(4(y^2 - 10y)\) is rewritten by completing to \(4((y-5)^2 - 25)\).

This allows us to easily transition into the standard form, making subsequent steps like finding vertices and foci straightforward.

Standard Form of Hyperbola

The standard form of a hyperbola provides a clear structure to identify its key components such as center, axis, vertices, and asymptotes.
The general standard form for a vertical-opening hyperbola is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), where \((h, k)\) is the center. Depending on whether the \(x\) or \(y\) term comes first, the hyperbola will open horizontally or vertically, respectively.

• After completing the square in this exercise, the hyperbola's standard form is \( \frac{(y-5)^2}{11} - \frac{(x+2)^2}{44} = 1 \), indicating it opens along the y-axis.

This form enables us to extract vital information like the location of the center \((-2, 5)\), and identify opening direction, making sketching and analysis much easier.