Problem 27 - Calculus Early Transcendentals

Question

Identify the type of conic section whose equation is given and find the vertices and foci. $$x^{2}=4 y-2 y^{2}$$

Step-by-Step Solution

Answer

The conic is a hyperbola with vertices at (0,0) and (0,2) and foci at \((0, 1\pm \frac{\sqrt{6}}{2})\)."

Step 1: Recognize the conic section

Rearrange the given equation: \( x^2 = 4y - 2y^2 \) into the standard form of a conic section. Begin by reordering terms to \( 2y^2 - 4y + x^2 = 0 \). This form suggests that this equation may define a conic section related to a parabola, ellipse, or hyperbola.

Step 2: Complete the square for the y-terms

Rearrange the equation to isolate the y-terms on one side: \( 2(y^2 - 2y) = -x^2 \). To complete the square, take the coefficient of \( y \) which is -2, halve it to -1, and square it to get 1. Add and subtract this inside the bracket: \( 2((y-1)^2 - 1) = -x^2 \). Simplify to get \( 2(y-1)^2 - 2 = -x^2 \).

Step 3: Simplify the equation

Rewrite the equation by moving the constant term to the other side: \( 2(y-1)^2 = -x^2 + 2 \). Simplify further to get \( (y-1)^2 = \frac{-1}{2}x^2 + 1 \). This cleaned up form indicates the conic is a hyperbola.

Step 4: Analyze the conic section's properties

The equation \( (y-1)^2 = \frac{-1}{2}x^2 + 1 \) is in the form of a vertical hyperbola centered at (0,1) with an opening parallel to the y-axis. Identify the nature of the conic as a hyperbola.

Step 5: Find the vertices and foci of the hyperbola

The standard form of a hyperbola is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Comparing, here \( a^2 = 1 \) and \( b^2 = \frac{1}{2} \), so vertices are at \((0,1 \pm a)\), i.e., \((0,0)\) and \((0,2)\). The distance from center to each focus is \( c = \sqrt{a^2 + b^2} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}\), so foci are at \((0, 1 \pm \frac{\sqrt{6}}{2})\).

Key Concepts

Hyperbolas

Hyperbolas are fascinating conics formed by the intersection of a double cone and a plane, creating two separate curves. A hyperbola is represented in a standard equation form like this:

• Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Where \((h, k)\) is the center of the hyperbola, \(a\) and \(b\) are constants that define its shape.
In our exercise, the given equation \((y-1)^2 = \frac{-1}{2}x^2 + 1\) shows a vertical hyperbola that opens upwards and downwards around its center \((0, 1)\). This form makes it clear that the y-axis plays a central role in shaping this hyperbola.
To identify and analyze a hyperbola, remember that the sign determines the direction it opens. In our case, since the y-term is positive, it opens vertically with respect to the y-axis.

Vertices and Foci

Vertices and foci are essential points that help shape and define hyperbolas.
The vertices are located at the shortest distance from the center to the curves, while the foci are positioned further out, providing a crucial aspect of the hyperbola's geometry.To find these points in a hyperbola:

• For vertices: The distance between the center \((0, 1)\) and the vertices is indicated by \( a \). For our vertical hyperbola, vertices can be found at \((0, 1\pm a)\), giving us \((0, 0)\) and \((0, 2)\).
• For foci: The distance to the foci involves \( c \), calculated using the relationship \( c = \sqrt{a^2 + b^2} \). Here, that becomes \( c = \frac{\sqrt{6}}{2} \), so foci are located at \((0, 1\pm \frac{\sqrt{6}}{2})\).

The vertices and foci of hyperbolas offer a vivid depiction of how the curves stretch and expand, highlighting their bounds around the center and demonstrating unique geometric properties.

Completing the Square

Completing the square is a powerful algebraic method used to convert quadratic equations into a form that reveals the nature of conic sections.
This process is particularly useful for rearranging and identifying hyperbolas, like in our example.The method involves transforming the y-terms \( 2(y^2 - 2y) = -x^2 \) by:

• Taking half of the linear y-term coefficient \(-2\), squaring it to get \(1\), and using it to complete the square: \(2((y-1)^2 - 1) = -x^2\).
• Reformulating yields \(2(y-1)^2 - 2 = -x^2\), thus isolating and tidying up the y-side of the equation.

Now, you can perceive the equation as a hyperbola, evident from the form \((y-1)^2 = \frac{-1}{2}x^2 + 1\).This clarity, provided by completing the square, enables a direct comparison with a standard hyperbola form, allowing you to timely find and interpret its key features, including vertices and foci in the resulting equation.