Problem 12 - Precalculus with Limits

Question

In Exercises 9-22, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. \(\dfrac{x^2}{36}-\dfrac{y^2}{4}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at origin (0,0). Vertices are located at (-6,0) and (6,0). Foci are at \((-2\sqrt{10},0)\) and \((2\sqrt{10},0)\). Asymptotes are given by the equations \(y = \pm \dfrac{1}{3}x\). The graph of the hyperbola opens along the x-axis with the branches bending towards the asymptotes of slope \(\pm \dfrac{1}{3}\).

Step 1: Find the Center

The center of the hyperbola is given by (h,k). Since there are no constants added to x or y in, h and k are both 0. So, the center of the hyperbola is at the origin (0,0).

Step 2: Find the Vertices

The vertices are a distance 'a' from the center along the x-axis in either direction. Since the value of 'a' is 6, the vertices are at points (-6,0) and (6,0).

Step 3: Find the Foci

The foci are found using the equation \(c = \sqrt{a^2 + b^2}\). Substituting the given values gives us \(c = \sqrt{6^2 + 2^2} = 2\sqrt{10}\). This means the foci are \(2\sqrt{10}\) units from the centre in either direction along the x-axis. Therefore, the foci are located at \((-2\sqrt{10},0)\) and \((2\sqrt{10},0)\).

Step 4: Determine the Asymptotes

The equations of the asymptotes for a hyperbola are \(y = \pm (b/a)x\). So, substituting our values for 'a' and 'b' gives us the asymptotes of the hyperbola as \(y = \pm \dfrac{2}{6}x\), which simplifies to \(y = \pm \dfrac{1}{3}x\). The asymptotes intersect at the center of the hyperbola and have slopes of \(\pm \dfrac{1}{3}\).

Step 5: Graph the Hyperbola

The hyperbola is now ready to sketch. Using the center, vertices, and foci helps to define the basic shape of the hyperbola. Draw the asymptote lines using their equations \(y = \pm \dfrac{1}{3}x\), and sketch the hyperbola opening towards the x-axis. The branch of the hyperbola nearest to the focus \(F_1\) lies entirely to the left of the vertical line through \(F_1\), and the branch nearest to the focus \(F_2\) lies entirely to the right of the vertical line through \(F_2\) .

Key Concepts

Center of Hyperbola

The center of a hyperbola is a critical point that acts as the reference for other elements like vertices and foci. A hyperbola in standard form either vertically or horizontally has a center at point (h, k). This can be determined from the equation by locating the constants subtracted from the x or y terms.

In the equation \(\dfrac{x^2}{36}-\dfrac{y^2}{4}=1\), there are no additional constants added to x or y. Thus, the center is at

• \((h, k) = (0, 0)\)

This center serves as the anchor point for sketching other features of the hyperbola like vertices, foci, and asymptotes, ensuring symmetry in the graph of the hyperbola.

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. These are essential in defining the shape and stretch of the hyperbola. For a horizontal hyperbola, the vertices are positioned along the x-axis.

From the hyperbola equation \(\dfrac{x^2}{36}-\dfrac{y^2}{4}=1\):

• 'a' value is derived from the denominator under \(x^2\); thus, \(a = \sqrt{36} = 6\).
• The vertices are located \(a\) units away from the center (0,0) along the x-axis.
• This makes the vertices at points \((-6,0)\) and \((6,0)\).

Plotting these vertices provides a visual range of how wide the branches of the hyperbola will be.

Foci of Hyperbola

The foci are integral for defining a hyperbola. They are located along the transverse axis and determine how the curve opens.

To find the foci, we use the relationship:

• \(c^2 = a^2 + b^2\)
• For our hyperbola equation: \(a^2 = 36\) and \(b^2 = 4\).
• This gives \(c = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\).

The foci are positioned \(c\) units away from the center along the x-axis:

• \((-2\sqrt{10}, 0)\) and \((2\sqrt{10}, 0)\)

Knowing the foci is crucial for understanding the direction and spread of the hyperbola.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches. They guide the shape and direction of the hyperbola's branches.

The equations of the asymptotes for a hyperbola help in sketching the curve accurately. For the given hyperbola \(\dfrac{x^2}{36} - \dfrac{y^2}{4} = 1\), the asymptote equations are derived as:

• \(y = \pm\dfrac{b}{a}x\) where \(a = 6\) and \(b = 2\).
• So the slopes are \(\pm \dfrac{2}{6} = \pm \dfrac{1}{3}\).
• This makes the equations \(y = \pm \dfrac{1}{3}x\).

These lines cross at the center (0,0) and provide an essential framework, allowing the correct shape of the hyperbola to be drawn.