Problem 29 - Calculus Early Transcendentals

Question

Identify the type of conic section whose equation is given and find the vertices and foci. $$y^{2}+2 y=4 x^{2}+3$$

Step-by-Step Solution

Answer

The conic is a hyperbola with vertices at (0, 1) and (0, -3); foci at (0, -1+√5) and (0, -1-√5).

Step 1: Rewrite the Equation

Begin by rewriting the given equation \(y^2 + 2y = 4x^2 + 3\). To better understand the conic, rearrange the terms involving each variable:\[ y^2 + 2y - 4x^2 = 3 \]

Step 2: Complete the Square for y

The term with \(y\) is in the form \(y^2 + 2y\). To complete the square, add and subtract \(1\) (since \((1/2) \times 2 = 1\), and \(1^2 = 1\)) to get:\[ (y^2 + 2y + 1) - 1 - 4x^2 = 3 \] This simplifies to:\[ (y+1)^2 - 4x^2 = 4 \]

Step 3: Recognize the Conic Section

Compare the equation \((y+1)^2 - 4x^2 = 4\) to the standard form of a hyperbola \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Divide each side by 4 to match this form:\[ \frac{(y+1)^2}{4} - x^2 = 1 \] This is the equation of a hyperbola centered at \((0, -1)\) and oriented vertically.

Step 4: Find Vertices

For a vertically oriented hyperbola defined by \(\frac{(y+1)^2}{4} - x^2 = 1\), the vertices are found by setting \(x = 0\) and solving \((y+1)^2 = 4\). Solving gives \[ y+1 = \pm 2 \Rightarrow y = 1 \text{ and } y = -3 \] Thus, the vertices are \((0, 1)\) and \((0, -3)\).

Step 5: Find Foci

For the hyperbola \(\frac{(y+1)^2}{4} - x^2 = 1\), use the relationship \(c^2 = a^2 + b^2\) where \(a^2 = 4\) and \(b^2 = 1\). Calculate:\[ c^2 = 4 + 1 = 5 \Rightarrow c = \sqrt{5} \]The foci are positioned along the y-axis, at \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\).

Step 6: Summarize Findings

The equation \(y^2 + 2y = 4x^2 + 3\) describes a hyperbola centered at \((0, -1)\) with vertices at \((0, 1)\) and \((0, -3)\), and foci at \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\).

Key Concepts

Hyperbola

A hyperbola is one of the familiar conic sections that you might come across during your mathematical journey. Conic sections include circles, ellipses, parabolas, and hyperbolas.
These are the curves obtained by intersecting a plane with a double-napped cone.
Among them, a hyperbola stands out due to its unique shape and properties.In the case of a hyperbola, you'll notice two separate, mirror-image curves. These curves are formed because the intersecting plane cuts through both nappes (top and bottom) of the cone.
What's essential about hyperbolas is their focus on two fixed points, called foci. The difference in the distances from any point on the hyperbola to these foci is always constant.
Hyperbolas are commonly represented by the standard equation:

• For vertically oriented hyperbolas: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
• For horizontally oriented hyperbolas: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)

These equations help us identify a hyperbola and locate its important features such as vertices and foci.

Completing the Square

Completing the square is a valuable algebraic technique for handling quadratic expressions.
It is especially useful when you're trying to transform an equation into a more recognizable standard form.
This technique helps to clarify the structure of the equation, particularly when working with conic sections.Consider the expression \(y^2 + 2y\), which appears in the equation \(y^2 + 2y = 4x^2 + 3\).
To complete the square for the \( y \) terms, you first identify the coefficient of \(y\), which is 2.
Taking half of this coefficient gives you 1, and squaring it gives 1. You then add and subtract this value within the equation:

• \(y^2 + 2y = (y^2 + 2y + 1) - 1\)

The expression \((y^2 + 2y + 1)\) becomes \((y+1)^2\), and the equation changes to

• \((y+1)^2 - 4x^2 = 4\)

By completing the square, this equation can be more easily identified as a hyperbola. Completing the square is an indispensable tool for furthering your understanding of conic sections.

Vertices and Foci

Finding the vertices and foci is essential when working with hyperbolas, as they reveal crucial details about the graph's orientation and dimensions.
Understanding these elements gives you insight into the hyperbola's geometric properties.To find the vertices and foci for a hyperbola in standard form, such as \(\frac{(y+1)^2}{4} - x^2 = 1\), you must identify the values of \(a^2\) and \(b^2\).

• Here, \(a^2 = 4\) and \(b^2 = 1\).

Vertices- The vertices are located \(a\) units from the center along the axis of the hyperbola.
For our equation, the center is at point \((0, -1)\), and \(a = 2\).
Thus, the vertices are at \((0, 1)\) and \((0, -3)\).Foci- The foci are located \(c\) units from the center, with the distance \(c\) calculated by \(c^2 = a^2 + b^2\).For our particular hyperbola,

• \(c^2 = 4 + 1 = 5\), so \(c = \sqrt{5}\).

This leads to the foci being positioned at \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\).
Understanding vertices and foci helps you sketch the hyperbola and understand its orientation and spread in the coordinate plane.