Problem 18 - 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. \(4 x^{2}-8 x-9 y^{2}-72 y+112=0\)

Step-by-Step Solution

Answer

The standard form is \(\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\); vertices at \((1 \pm 3\sqrt{7}, -4)\); foci at \((1 \pm \sqrt{91}, -4)\); asymptotes \(y = -4 \pm \frac{\sqrt{28}}{\sqrt{63}}(x-1)\).

Step 1: Organize the Equation

Start by rewriting the given equation, keeping the terms involving \(x\) and \(y\) together: \[4x^2 - 8x - 9y^2 - 72y = -112\] Now, rearrange to group \(x\) and \(y\) terms separately:\[4(x^2 - 2x) - 9(y^2 + 8y) = -112\]

Step 2: Complete the Square

Complete the square for the \(x\) terms and \(y\) terms separately:- For \(x\): \[x^2 - 2x = (x-1)^2 - 1\]- For \(y\): \[y^2 + 8y = (y+4)^2 - 16\]Substitute these into the equation:\[4((x-1)^2 - 1) - 9((y+4)^2 - 16) = -112\]

Step 3: Simplify the Equation

Simplify the equation by distributing the coefficients and combining like terms:\[4(x-1)^2 - 4 - 9(y+4)^2 + 144 = -112\]Combine constants on the right side:\[4(x-1)^2 - 9(y+4)^2 = -112 + 4 - 144 = -252\]

Step 4: Divide by the Constant

Divide the entire equation by \(-252\) to set the right side equal to 1:\[\frac{4(x-1)^2}{-252} - \frac{9(y+4)^2}{-252} = 1\]Simplify to:\[\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\]

Step 5: Identify Key Components

Based on the standard form of the hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), identify:- Center \((h, k) = (1, -4)\)- \(a^2 = 63\), \(b^2 = 28\)Calculate \(a = \sqrt{63}\) and \(b = \sqrt{28}\).

Step 6: Find Vertices and Foci

Vertices are found using \((h \pm a, k)\): \( (1 \pm \sqrt{63}, -4) = (1 \pm 3\sqrt{7}, -4)\).Calculate \(c = \sqrt{a^2 + b^2} = \sqrt{63 + 28} = \sqrt{91}\).Foci \( (1 \pm \sqrt{91}, -4)\).

Step 7: Write Equations of Asymptotes

The asymptotes are derived from the formula \(y = k \pm \frac{b}{a}(x - h)\).Substitute the known values:\(y = -4 \pm \frac{\sqrt{28}}{\sqrt{63}}(x - 1)\).Simplify the asymptote equations if needed.

Key Concepts

Vertices

The vertices of a hyperbola are crucial points where the curve reaches its widest part. They lie on the transverse axis, which is the line that intersects the hyperbola at these points. In standard form, the equation of a hyperbola is typically written as \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] or its vertical orientation equivalent. From this form,

• The vertices can be found using the formula \((h \pm a, k)\), if the hyperbola opens horizontally, or \((h, k \pm a)\), if it opens vertically.
• For the given problem, with the simplified equation \[\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\], we have the vertices at \((1 \pm 3\sqrt{7}, -4)\), indicating that this hyperbola opens horizontally.

The value \(a\ = \sqrt{63}\) provides the distance from the center to each vertex along the horizontal axis.

Foci

The foci of a hyperbola are points within each arm of the hyperbola, aligned along the transverse axis. These points are significant because they help define the hyperbola's shape. You can use the relationship:

• \(c^2 = a^2 + b^2\)

This relationship is essential for identifying how far each focus is from the center of the hyperbola. For a hyperbola in standard form, the coordinates of the foci are

• For a horizontal hyperbola: \((h \pm c, k)\)
• For a vertical hyperbola: \((h, k \pm c)\)

In the current problem, we calculated \(c = \sqrt{91}\), which gives us the foci at \((1 \pm \sqrt{91}, -4)\), lying along the same horizontal line as the vertices.

Equations of Asymptotes

Asymptotes are the invisible lines that a hyperbola approaches but never touches. They guide the overall behavior of the hyperbola. For hyperbolas, the equations of the asymptotes are derived from:\[ y = k \pm \frac{b}{a}(x-h) \]This formula comes into play when the hyperbola is expressed in standard form. Depending on whether the hyperbola opens horizontally or vertically, substituting the known values will give you the asymptotic directions in the Cartesian plane.In the provided exercise:

• With \(b = \sqrt{28} \) and \(a = \sqrt{63}\), we substitute into the formula for the asymptotes: \[y = -4 \pm \frac{\sqrt{28}}{\sqrt{63}}(x - 1)\].
• This simplifies to the asymptotic directions, indicating how the arms of the hyperbola spread away infinitely.

Standard Form

The standard form of a hyperbola's equation is a structured way of representing these curves, which allows easy identification of the center, vertices, and orientation. A hyperbola's standard form can be written as:

• \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] for hyperbolas opening horizontally.
• \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\] for hyperbolas opening vertically.

The standard form is crucial because it allows:

• Calculation of the vertices using \(a\),
• Determination of the distance to the foci using \(c\),
• Identification of asymptotes through simple formulas.

In the initial problem, transforming \[4x^2 - 8x - 9y^2 - 72y = -112\] into \[\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\] not only reveals the center \((1, -4)\) but also the \(a^2=63\) and \(b^2=28\) values needed for further calculations.