Problem 10 - 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. \(\dfrac{x^2}{9}-\dfrac{y^2}{25}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at (0,0), vertices are located at (3,0) and (-3,0), foci at (\sqrt{34}, 0) and (-\sqrt{34}, 0), equations of the asymptotes are \(y = \frac{5}{3}x\) and \(y = -\frac{5}{3}x\).

Step 1: Identify the Center

The center of the hyperbola can be found at \((h, k)\) which in this case is \((0, 0)\) as there are no additions or subtractions in our equation.

Step 2: Find the Vertices

The vertices of the hyperbola are at \((h + a, k)\) and \((h - a, k)\) in the case of a horizontal transverse axis (as in our case). In our exercise, since \(a = \sqrt{9} = 3\), we get the vertices are at \((0 + 3, 0) = (3, 0)\) and \((0 - 3, 0) = (-3, 0)\).

Step 3: Find the Foci

The foci of the hyperbola are at \((h + c, k)\) and \((h - c, k)\) in case of a horizontal transverse axis. Here, to find the distance to the foci \(c\), we use the formula \(c = \sqrt{a^2 + b^2} = \sqrt{9 + 25} = \sqrt{34}\). So, the foci are located at \((0 + \sqrt{34}, 0) = (\sqrt{34}, 0)\) and \((0 - \sqrt{34}, 0) = (-\sqrt{34}, 0)\).

Step 4: Asymptotes Equation

The equations of the asymptotes of a hyperbola are given by \(y = k + \frac{b}{a}(x - h)\) for the upper branch and \(y = k - \frac{b}{a}(x - h)\) for the lower branch. In this case, the center is \((0,0)\) and \(a = 3\), \(b = 5\). Substituting these values we get \(y = \frac{5}{3}x\) and \(y = -\frac{5}{3}x\).

Step 5: Sketch the Graph

Now that we have the center, vertices, foci, and asymptotes of the hyperbola, we can sketch the graph. Since it's a symmetrical to x-axis hyperbola, there will be two branches. The center, vertices, foci will guide the placement and shape, and the asymptotes will guide the direction of the branches.

Key Concepts

Understanding Conic Sections

Hyperbolas are part of a larger family called conic sections. Conic sections are curves obtained by slicing a cone with a plane. They come in different types, including circles, ellipses, parabolas, and hyperbolas, depending on the angle at which the plane intersects the cone.
For hyperbolas, the plane cuts through both halves of the cone, producing two separate curves. Each part of a hyperbola mirrors the other, and they open away from each other. This unique shape is what differentiates hyperbolas from other conic sections.

• Circle: Plane cuts parallel to the base.
• Ellipse: Plane cuts at an angle, but not steep enough.
• Parabola: Plane cuts parallel to a generatrix.
• Hyperbola: Plane cuts through both parts of double cone.

Recognizing these distinct features helps in understanding hyperbolic shapes and their properties.

Asymptotes and Their Role

Asymptotes are crucial in understanding the behavior of hyperbolas. These are straight lines that the hyperbola approaches but never touches. They guide us in sketching the curve accurately.
For hyperbolas defined by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations of the asymptotes are \( y = \pm \frac{b}{a}x \). In our case:

• Upper Branch: \( y = \frac{5}{3}x \)
• Lower Branch: \( y = -\frac{5}{3}x \)

Knowing the asymptotes helps us draw initial guidelines on the graph, ensuring that the hyperbola extends along these directions, creating a precise parabolic shape.

Vertices: The Starting Points

The vertices of a hyperbola are key points where the curve is closest to the center. They define the width of the opening between the branches.
For a hyperbola with a horizontal transverse axis like ours, the formula for the vertices is:\((h \pm a, k)\). Here, \(a\) is derived from \(a^2 = 9\), making \(a = 3\). Thus, our vertices are:

• \((3, 0)\)
• \((-3, 0)\)

The vertices not only help in drawing the hyperbola but also establish the baseline from which the branches diverge, giving the curve structure and form.

Foci: The Hidden Centers

Foci are the fixed points crucial in defining a hyperbola. The distance of each point on a hyperbola from the foci follows a specific relationship.
The formula for finding the foci in horizontal hyperbolas like ours is \((h \pm c, k)\) where \(c = \sqrt{a^2 + b^2}\). With \(a^2 = 9\) and \(b^2 = 25\), we calculate \(c = \sqrt{34}\). So, the foci lie at:

• \((\sqrt{34}, 0)\)
• \((-\sqrt{34}, 0)\)

The positioning of the foci influences the curve's openness, making them indispensable for accurately plotting and understanding hyperbolas.