Problem 7 - 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. $$ 5 x^{2}-4 y^{2}-40 x-16 y-36=0 $$

Step-by-Step Solution

Answer

Center: (4, -2). Vertices: (4, 3) and (4, -7). Foci: (4±3√5, -2). Asymptotes: y+2=±√5/2(x-4).

Step 1: Rearrange the Equation

Start by rewriting the given equation of the hyperbola in a form that helps us complete the square. Rearrange the equation as:\[ 5x^2 - 40x - 4y^2 - 16y = 36 \]

Step 2: Complete the Square for x-term

Factor out the coefficient of the \(x\) terms:\[ 5(x^2 - 8x) - 4y^2 - 16y = 36 \]Complete the square inside the parentheses for \(x\): - Take half of \(-8\), which is \(-4\), and square it, resulting in \(16\).Add and subtract \(16\) inside the parentheses:\[ 5((x^2 - 8x + 16) - 16) - 4y^2 - 16y = 36 \]Rearrange to:\[ 5((x-4)^2 - 16) - 4y^2 - 16y = 36 \]Then distribute \(5\):\[ 5(x-4)^2 - 80 - 4y^2 - 16y = 36 \]

Step 3: Complete the Square for y-term

Next, handle \(y\) terms by factoring and completing the square:Factor \(-4\) from the \(y\) terms:\[ 5(x-4)^2 - 4(y^2 + 4y) - 80 = 36 \]Complete the square for \(y\): - Take half of \(4\), which is \(2\), and square it, resulting in \(4\).Add and subtract \(4\) within the parentheses:\[ 5(x-4)^2 - 4((y^2 + 4y + 4) - 4) - 80 = 36 \]Rearrange to:\[ 5(x-4)^2 - 4(y+2)^2 + 16 - 80 = 36 \]

Step 4: Simplify and Write the Standard Equation

Combine and simplify constant terms on the left-hand side to solve for the standard form:\[ 5(x-4)^2 - 4(y+2)^2 - 64 = 36 \]Add \(64\) to both sides:\[ 5(x-4)^2 - 4(y+2)^2 = 100 \]Divide every term by \(100\) to get the equation in standard form:\[ \frac{(x-4)^2}{20} - \frac{(y+2)^2}{25} = 1 \]

Step 5: Identify the Hyperbola Features

The standard form equation \( \frac{(x-4)^2}{20} - \frac{(y+2)^2}{25} = 1 \) indicates a hyperbola centered at \((h,k) = (4, -2)\) with \(a^2 = 25\) and \(b^2 = 20\). Thus, \(a = 5\) and \(b = \sqrt{20} = 2\sqrt{5}\). Determine \(c\) using \(c^2 = a^2 + b^2\):\[ c^2 = 25 + 20 = 45 \]So, \(c = \sqrt{45} = 3\sqrt{5}\). Thus, the foci are at \((4 \pm 3\sqrt{5}, -2)\).

Step 6: Determine Asymptotes Equations

For hyperbolas of the form \( \frac{(x-h)^2}{b^2} - \frac{(y-k)^2}{a^2} = 1 \), asymptotes are:\[ y-k = \pm \frac{a}{b} (x-h) \]Substitute the values of \((h,k), a, b\):\[ y + 2 = \pm \frac{5}{2\sqrt{5}}(x-4) \] Rationalizing, \( \frac{5}{2\sqrt{5}} = \frac{5\sqrt{5}}{10} = \frac{\sqrt{5}}{2} \)Asymptote equations are:\[ y + 2 = \pm \frac{\sqrt{5}}{2}(x-4) \]Thus, rearranging, we have:\[ y = \pm \frac{\sqrt{5}}{2}(x-4) - 2 \]

Step 7: List the Coordinates of Vertices and Graph

The vertices are located at \((h, k \pm a)\), giving us the coordinates:\( (4, -2\pm 5) = (4, 3) \text{ and } (4, -7) \).Graph the hyperbola using these vertices, center, foci, and asymptotes for guidance.

Key Concepts

Coordinates of Vertices

Understanding the vertices of a hyperbola is crucial, as they help shape its graph.
In hyperbolas, the vertices are points where the hyperbola intersects its main axis. For the given hyperbola equation in standard form: \[ \frac{(x-4)^2}{20} - \frac{(y+2)^2}{25} = 1 \]we identify the vertices using the center and the value of \(a\) from the equation.
The center

• is \((h, k) = (4, -2)\)

The given equation indicates that \(a^2 = 25\), so \(a = 5\).
This means the vertices, which are found at \((h, k \pm a)\), are located at:

• \((4, -2 + 5) = (4, 3)\)
• \((4, -2 - 5) = (4, -7)\)

These vertices help define the shape and orientation of the hyperbola along its vertical axis.

Equations of Asymptotes

Asymptotes are lines that the branches of a hyperbola approach but never touch. They give a visual guide to the hyperbola's direction. For a hyperbola in the form: \[\frac{(x-h)^2}{b^2} - \frac{(y-k)^2}{a^2} = 1\]the equations for the asymptotes are derived from the slopes \(\pm \frac{a}{b}\):- Asymptote equation is given by:\[ y - k = \pm \frac{a}{b}(x - h) \]With known values:

• \((h, k) = (4, -2)\)
• \(a = 5\)
• \(b = \sqrt{20} = 2\sqrt{5}\)

The slope \(\frac{a}{b}\) is simplified from \(\frac{5}{2\sqrt{5}}\) to \(\frac{\sqrt{5}}{2}\), by rationalizing.
Thus, the asymptotes are described by:

• \[ y + 2 = \pm \frac{\sqrt{5}}{2} (x - 4) \]
• which rearranges to \[ y = \pm \frac{\sqrt{5}}{2} (x - 4) - 2 \]

These equations reveal the lines across which the hyperbola stretches infinitely.

Standard Form of Hyperbola

The standard form of a hyperbola gives an organized way to interpret its features and sketch its graph. For hyperbolas, the standard equation is either:

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

This tells us how the hyperbola opens—either horizontally or vertically.
For the equation we derived:\[ \frac{(x-4)^2}{20} - \frac{(y+2)^2}{25} = 1 \]it shows that the hyperbola opens along the \(y\)-axis because the \(y\)-term has a larger denominator.
Key parameters include:

• Center \((h,k) = (4,-2)\)
• \(a = 5\)
• \(b = 2\sqrt{5}\)

These values help us sketch the hyperbola by marking key points and asymptotes, giving a clear view of its structure.

Completing the Square

Completing the square is a technique used to simplify and solve quadratic equations, making it easier to visualize conic sections like hyperbolas. It involves rearranging and balancing parts of an equation to form perfect square trinomials.
Let's see how it was applied to the exercise:Step-by-Step Approach:

• Identify terms: Start with grouped \(x\) and \(y\) parts: \(5(x^2 - 8x) - 4(y^2 + 4y)\).
• Complete squares: For \(x\), half of \(-8\) is \(-4\), square this to get \(16\), and add/subtract it within the group.
• For \(y\), half of \(4\) is \(2\), square to get \(4\), perform similar adjustments.

After these adjustments and simplifications:

• \[ 5((x-4)^2 - 16) - 4((y+2)^2 - 4) = 36 \]
• Convert to standard form: \[ \frac{(x-4)^2}{20} - \frac{(y+2)^2}{25} = 1 \]

Completing the square method makes identifying the hyperbola's characteristics straightforward.