Problem 68 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ y^{2}-4 x^{2}+16 x=24 $$

Step-by-Step Solution

Answer

Center: (2, 0); Vertices: (2, ±2√2); Foci: (2, ±√10); Asymptotes: y=2(x-2), y=-2(x-2).

Step 1: Rewrite the Equation in Standard Form

To find the characteristics of the conic, we first transform the given equation into its standard form.Start with the equation: \[ y^{2} - 4x^{2} + 16x = 24 \]Rearrange terms related to \(x\): \[ y^{2} - 4(x^{2} - 4x) = 24 \] Complete the square for the expression in parentheses:1. Take the coefficient of \(x\), which is \(-4\) in \(-4x\).2. Divide by 2 to get \(-2\).3. Square the result to get \(4\). Add and subtract this value inside the parentheses: \[ y^{2} - 4(x^{2} - 4x + 4 - 4) = 24 \] This simplifies to: \[ y^{2} - 4((x-2)^2 - 4) = 24 \] Simplify further:\[ y^{2} - 4(x-2)^{2} + 16 = 24 \]Subtract 16 from both sides:\[ y^{2} - 4(x-2)^{2} = 8 \] Divide the entire equation by 8 to get the form for a hyperbola:\[ \frac{y^{2}}{8} - \frac{(x-2)^{2}}{2} = 1 \]

Step 2: Identify the Center of the Hyperbola

The standard form for a hyperbola is \(\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1\), where the center is \((h, k)\).In our equation, we have: \[ \frac{y^{2}}{8} - \frac{(x-2)^{2}}{2} = 1 \]Thus, the center is at \((h, k) = (2, 0)\).

Step 3: Find the Vertices

In a hyperbola of this form, the vertices occur along the 'positive' part of the standard hyperbola equation. Here, \(y^{2}\) has the larger denominator, so the hyperbola opens up and down.The distance \(a\) from the center to each vertex along the y-axis is \(\sqrt{8} = 2\sqrt{2}\).So, the vertices are at:\[(h, k \pm a) = (2, 0 \pm 2\sqrt{2}) \Rightarrow (2, 2\sqrt{2}), (2, -2\sqrt{2})\].

Step 4: Identify the Foci

The distance from the center to each focus is given by \(c\), where \(c^2 = a^2 + b^2\).Here, \(a^2 = 8\) and \(b^2 = 2\).Therefore, \(c^2 = 8 + 2 = 10\) giving \(c = \sqrt{10}\).So, the foci are at:\[(h, k \pm c) = (2, 0 \pm \sqrt{10}) \Rightarrow (2, \sqrt{10}), (2, -\sqrt{10})\].

Step 5: Determine the Asymptotes

For a hyperbola of the form \(\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1\), the asymptotes are given by the lines:\[y - k = \pm \frac{a}{b} (x - h)\]Here, \(a = \sqrt{8}\) and \(b = \sqrt{2}\).The slopes of the asymptotes are \(\pm \frac{\sqrt{8}}{\sqrt{2}} = \pm \sqrt{4} = \pm 2\).Thus, the equations of the asymptotes are:\[y = 2(x - 2)\] and \[y = -2(x - 2)\].

Key Concepts

Hyperbola

A hyperbola is a type of conic section that is formed by intersecting a double cone with a plane in such a way that the angle between the plane and the cone’s base is less than the cone’s opening angle, but not parallel to the axis of the cones. Unlike an ellipse or a circle, a hyperbola consists of two distinct, separate curves known as branches. Each branch appears as a curved line that extends outward indefinitely. The standard mathematical representation of a hyperbola depends on which direction it opens. For hyperbolas that open up and down or left and right, the general form involves squared terms with opposite signs, such as

• \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \) (opening up and down)
• \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) (opening left and right)

Understanding hyperbolas is crucial because they appear in various fields, from physics to astronomy, where the trajectory of objects follows hyperbolic paths.

Standard Form

The standard form of the hyperbola equation is particularly helpful to identify its properties, including the center, axes, vertices, and foci. Transform the given equation into its standard form to easily derive all these characteristics. To convert an equation like \( y^{2} - 4x^{2} + 16x = 24 \) into its standard form, complete the square for both variables if necessary and ensure the equation equals 1. Typically, this includes rearranging terms and possibly dividing every term by a common factor.For example, after completion of the square and simplification, you might arrive at an equation like\( \frac{y^{2}}{8} - \frac{(x-2)^{2}}{2} = 1 \). Here, the equation's appearance matches the standard form enabling a straightforward analysis of its geometric properties.

Asymptotes

Asymptotes of a hyperbola are the lines that the branches of the hyperbola approach but never actually touch. They essentially define the "direction" the hyperbola opens out to infinity. For hyperbolas expressed in the standard form \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), the equations of the asymptotes can be derived. The general method involves using the formula:

• \( y - k = \pm \frac{a}{b} (x - h) \)

Given in our example, the slope calculated as \( \pm 2 \), leads to the asymptote equations: \( y = 2(x - 2) \) and \( y = -2(x - 2) \). These asymptotic lines help in sketching the hyperbola on a graph, indicating how the branches will curve and extend.

Vertices

Vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. They can be easily determined from the hyperbola’s standard form once you have carefully rewritten the equation. For example, if the hyperbola's equation is such that it opens along the y-axis, the vertices' coordinates are \( (h, k \pm a) \).Using the information derived from comparing the hyperbola's equation to its standard form, \( \frac{y^{2}}{8} - \frac{(x-2)^{2}}{2} = 1 \), the calculated distance 'a' from the center to each vertex is \( \sqrt{8} \). Therefore, the hyperbola's vertices are \( (2, 0 \pm 2\sqrt{2}) \), resulting in the points \( (2, 2\sqrt{2}) \) and \( (2, -2\sqrt{2}) \). These vertices help in establishing the overall shape and orientation of the hyperbola.

Foci

The foci of a hyperbola are crucial points located outside each branch of the hyperbola, lying along the transverse axis. They play a significant role in defining and understanding the shape, as the difference of the distances from any point on the hyperbola to the two foci is constant. For hyperbolas in standard form, the foci can be calculated using the relationship in the hyperbola's geometry. If you have \( c^{2} = a^{2} + b^{2} \), then \( c \) is the distance from the center to each focus.For the equation \( \frac{y^{2}}{8} - \frac{(x-2)^{2}}{2} = 1 \), given \( a^{2} = 8 \) and \( b^{2} = 2 \), we find \( c^{2} = 10 \) and consequently \( c = \sqrt{10} \). Thus, the foci are found at \( (2, 0 \pm \sqrt{10}) \), or \( (2, \sqrt{10}) \) and \( (2, -\sqrt{10}) \). Identifying the foci helps in accurately plotting the hyperbola and understanding its geometric properties.