Problem 23 - Calculus Early Transcendentals

Question

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$4 x^{2}-y^{2}-24 x-4 y+28=0$$

Step-by-Step Solution

Answer

Vertices: (4, -2), (2, -2); Foci: (3±√5, -2); Asymptotes: y=2x-8, y=-2x+4.

Step 1: Rewrite the Equation

Start by rewriting the given hyperbola equation in a recognizable standard form. The equation is \(4x^2 - y^2 - 24x - 4y + 28 = 0\). First, we group the \(x\) and \(y\) terms as follows: \(4(x^2 - 6x) - (y^2 + 4y) = -28\).

Step 2: Complete the Square

Complete the square for the \(x\) and \(y\) terms separately. For the \(x\) terms: \(x^2 - 6x\) becomes \((x-3)^2 - 9\). For the \(y\) terms: \(y^2 + 4y\) becomes \((y+2)^2 - 4\). Substitute back into the equation: \(4((x-3)^2 - 9) - ((y+2)^2 - 4) = -28\).

Step 3: Simplify the Expression

Distribute and simplify: \(4(x-3)^2 - 36 - (y+2)^2 + 4 = -28\). Then, combine and rearrange: \(4(x-3)^2 - (y+2)^2 = 4\). Divide throughout by 4 to get \((x-3)^2 - \frac{(y+2)^2}{4} = 1\). Now, the equation is in standard form of a hyperbola.

Step 4: Identify Key Parameters

The standard form of a hyperbola is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). From our equation \((x-3)^2 - \frac{(y+2)^2}{4} = 1\), identify \(h = 3\), \(k = -2\), \(a^2 = 1\), and \(b^2 = 4\).

Step 5: Calculate Vertices

For hyperbolas, vertices are at \((h \pm a, k)\) for this orientation. Here, \(a = 1\), giving vertices at \((3 \pm 1, -2)\) which are \((4, -2)\) and \((2, -2)\).

Step 6: Calculate Foci

For foci, use \(c = \sqrt{a^2 + b^2} = \sqrt{1 + 4} = \sqrt{5}\). Foci are at \((h \pm c, k)\) which are at \((3 \pm \sqrt{5}, -2)\). Thus, the foci are at \((3 + \sqrt{5}, -2)\) and \((3 - \sqrt{5}, -2)\).

Step 7: Determine Asymptotes

The equations of the asymptotes for hyperbolas in this orientation are \(y = k \pm \frac{b}{a}(x-h)\). Here, \(b/a = 2/1 = 2\). Thus, the asymptotes are \(y = -2 \pm 2(x-3)\). These simplify to \(y = 2x - 8\) and \(y = -2x + 4\).

Step 8: Sketch the Hyperbola

Draw a Cartesian plane, plot the center \((3, -2)\), and mark the vertices \((4, -2)\) and \((2, -2)\). Plot the foci at \((3 + \sqrt{5}, -2)\) and \((3 - \sqrt{5}, -2)\), and draw the asymptotes \(y = 2x - 8\) and \(y = -2x + 4\) as dashed lines to guide the shape of the hyperbola.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its principal axis. In standard form, the equation of a hyperbola can guide us to these vertices easily.
The derived standard form from the given equation is \((x-3)^2 - \frac{(y+2)^2}{4} = 1\). Here, the center is at \((h, k) = (3, -2)\) and \(a^2 = 1\). Thus, \(a = 1\), indicating the distance from the center to each vertex.

• The vertices are identified based on the formula \((h \pm a, k)\).
• This gives us the vertices at \((3 \pm 1, -2)\), which simplifies to the points \((4, -2)\) and \((2, -2)\).

Understanding the vertices helps in grasping the basic structure of the hyperbola as they represent its closest approach to the center along the x-axis.

Foci of Hyperbola

In hyperbolas, the foci are key points that give the hyperbola its shape. Much like the vertices, the foci are symmetrically placed along the main axis of the hyperbola.
For the equation \((x-3)^2 - \frac{(y+2)^2}{4} = 1\), we need to calculate the distance to the foci using the formula \(c = \sqrt{a^2+b^2}\). Here, \(a^2 = 1\) and \(b^2 = 4\), giving \(c = \sqrt{5}\).

• The foci are positioned at \((h \pm c, k)\).
• Substituting the values, the foci are located at \((3 \pm \sqrt{5}, -2)\), leading to the points \((3 + \sqrt{5}, -2)\) and \((3 - \sqrt{5}, -2)\).

The foci are crucial for determining the opening and spread of the branches of a hyperbola.

Asymptotes of Hyperbola

Asymptotes are the diagonal lines that guide the shape of a hyperbola. They are what the branches of the hyperbola approach as they move further away from the center.
In our case, the standard form indicates the asymptotes using the relation \(y = k \pm \frac{b}{a}(x-h)\). With \(b = 2\) and \(a = 1\), we have \(b/a = 2\).

• The asymptotes thus become \(y = -2 \pm 2(x-3)\).
• This results in the equations \(y = 2x - 8\) and \(y = -2x + 4\).

These asymptotes define the boundary within which the branches of the hyperbola will exist, never actually touching but always approaching.