Problem 37 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and asymptotes of the hyperbola, and (c) sketch the hyperbola. Use a graphing utility to verify your graph. \(2 x^{2}-7 y^{2}+16 x+18=0\)

Step-by-Step Solution

Answer

The standard form is \(-(x+4)^2/1 + y^2/(2/7) = 1\). The center is at (-4,0), vertices are at (-5,0) and (-3,0), foci are at \((-4 - \sqrt{51}/7, 0)\) and \((-4 + \sqrt{51}/7, 0)\), and the asymptotes are \(y = 2/7 * (x + 4)\) and \(y = -2/7 *(x + 4)\). Use a graphing utility to verify these figures.

Step 1: Convert to Standard Form

To convert to standard form, we complete the squares for \(x\) and \(y\). First, rewrite the equation and group the \(x\)-terms and \(y\)-terms together: \(2(x^2 + 8x) - 7y^2 = -18\). Then, complete the square for the \(x\) terms by adding \((8/2)^2 = 16\) inside the bracket and to the other side of the equation to keep it balanced. Do same for \(y\), although in this case, it's already a perfect square so nothing is done: \(2(x+4)^2 - 7 y^2 = -2\). Divide through by -2 to to get standard form: \(-(x+4)^2/1 + y^2/(2/7) = 1\)

Step 2: Find the center, vertices, and foci

The center is \((-h, -k)\) which is \((-4, 0)\). The vertices are located at \(a\) units to the left and right of the center h, so they are at \((h-a, k)\) and \((h+a, k)\), which are \((-5, 0)\) and \((-3,0)\). To find the foci, we use the formula \(c = \sqrt{a^2 + b^2}\). In this case, \(c = \sqrt{1^2 + (2/7)^2} = \sqrt{51/49}\). Thus, the foci are at \((h-c, k)\) and \((h+c, k)\), which are \((-4 - \sqrt{51}/7, 0)\) and \((-4 + \sqrt{51}/7, 0)\).

Step 3: Calculate asymptotes and sketch the hyperbola

The equations for the asymptotes of a hyperbola centered at \((h, k)\) with stretch factors \(a\) and \(b\) are \(y = k + (x-h)*b/a\) and \(y = k - (x-h)*b/a\). So, plug \(h = -4\), \(k = 0\), \(a = 1\), \(b = 2/7\) into equations, we get the asymptote equations: \(y = 2/7 * (x + 4)\) and \(y = -2/7 *(x + 4)\). For sketching the hyperbola: draw center at (-4,0), plot vertices, draw asymptotes, and sketch in the two branches of the hyperbola.

Key Concepts

Standard Form of a Hyperbola

To transform any hyperbola equation into its standard form, we typically begin by rearranging and completing the square. This process is essential because it helps us identify the hyperbola's main geometric features such as the center, vertices, and foci. In the example provided, the equation is initially rewritten to show grouped terms of \(x\) and \(y\), i.e., \(2(x^2 + 8x) - 7y^2 = -18\).
Completing the square is performed only for the \(x\) terms by adding the square of half the coefficient of \(x\), which is \((8/2)^2 = 16\). This adjustment ensures the equation takes on the form \(-(x+4)^2/1 + y^2/(2/7) = 1\), a standard form for a horizontally oriented hyperbola
. The main goal here is to achieve a format where the expressions are equated to 1, aligning perfectly with the hyperbola equation \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\).
This makes it much easier to determine the hyperbola's properties.

Vertices of a Hyperbola

In a hyperbola, the vertices represent the points where the branches are closest together. Once the equation is in standard form, identifying the vertices involves looking at the values of \(a\) and \(b\). These govern how far the vertices are from the hyperbola's center.
For the equation \(-(x+4)^2/1 + y^2/(2/7) = 1\), it implies that \(a = 1\). The center of this hyperbola is \((-4, 0)\).
To find the vertices:\

• Add and subtract \(a\) from the \(x\)-coordinate of the center because the hyperbola opens horizontally.
• Vertices are therefore \((-4-1, 0)\) and \((-4+1, 0)\), resulting in coordinates \((-5, 0)\) and \((-3, 0)\).

Understanding the specific placement of vertices aids in drawing the correct shape of the hyperbola on a graph.

Foci of a Hyperbola

The foci of a hyperbola are critical points located along the transverse axis. They lie beyond each vertex and significantly influence the shape of the hyperbola. The distance of the foci from the center is defined by \(c\), where \(c = \sqrt{a^2 + b^2}\).
In our hyperbola equation \(-(x+4)^2/1 + y^2/(2/7) = 1\), the values are \(a = 1\) and \(b = \sqrt{2/7}\).
Calculating \(c\):

• Find \(c\) using \(c = \sqrt{1^2 + (\sqrt{2/7})^2} = \sqrt{51/49}\).
• The foci are positioned at \((h-c, k)\) and \((h+c, k)\), yielding coordinates ( \(-4-\sqrt{51}/7, 0\) and \(-4+\sqrt{51}/7, 0\)).

The foci help determine the hyperbola's eccentricity and verify that the branches diverge towards these focal points.

Asymptotes of a Hyperbola

Hyperbola asymptotes are imaginary lines that the branches approach but never actually meet. They are crucial guides for sketching hyperbolas as they reflect the general orientation and "spread" of the hyperbola's arms.
For the standard form \(-(x+4)^2/1 + y^2/(2/7) = 1\), asymptotes can be found using the equation \(y = k \pm \frac{b}{a}(x-h)\).
Here's how you derive the asymptotes for this hyperbola:

• Substitute the values \(h = -4\), \(k = 0\), \(a = 1\), and \(b = \sqrt{2/7}\) in the asymptote formula.
• This results in two asymptote equations: \(y = \frac{\sqrt{2/7}}{1}(x+4)\) and \(y = -\frac{\sqrt{2/7}}{1}(x+4)\).

These equations highlight the slope and direction of the asymptotes. Incorporating them into your graph ensures an accurate representation of the hyperbola’s shape.