Problem 63 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}+2 y^{2}-2 x-4 y=-1 $$

Step-by-Step Solution

Answer

The ellipse is centered at (1, 1) with vertices (1 ± √2, 1), and foci (0, 1) and (2, 1). It has no asymptotes.

Step 1: Rewrite the Equation

First, group the terms with the same variable together: \[ x^2 - 2x + 2y^2 - 4y = -1 \] and organize them in the standard form for conics. We will complete the square for both the \(x\)-terms and \(y\)-terms to facilitate this.

Step 2: Complete the Square for x-Terms

Focus on the \(x\)-terms: \[ x^2 - 2x \]To complete the square, take half of the coefficient of \(x\), square it, and add and subtract it inside the equation:\[ x^2 - 2x = (x-1)^2 - 1 \] Add \(1\) to \(-1\) on the right side to balance the equation.

Step 3: Complete the Square for y-Terms

Now focus on the \(y\)-terms: \[ 2(y^2 - 2y) \]Take half of \(2y\)'s coefficient \(\frac{-4}{2} = -2\), square it, and add and subtract its square inside the parentheses:\[ 2(y^2 - 2y) = 2((y-1)^2 - 1) = 2(y-1)^2 - 2 \]. Add 2 to the right side to balance the equation.

Step 4: Final Equation Transformation

After completing the square, rewrite the equation accounting for the terms you balanced:\[ (x-1)^2 + 2(y-1)^2 = 2 \] Divide through by 2 to form a standard ellipse equation:\[ \frac{(x-1)^2}{2} + \frac{(y-1)^2}{1} = 1 \]. Here, the equation represents an ellipse centered at (1, 1).

Step 5: Determine the Center and Vertices

The ellipse’s center is at \((h, k) = (1, 1)\). The vertices are found along the major axis (x-axis in this case): \[ (h \pm \, \sqrt{2}, k) = (1 \pm \sqrt{2}, 1) \].

Step 6: Compute Foci

To find the foci, use the formula \( c^2 = a^2 - b^2 \):\( b^2 = 1 \) and \(a^2 = 2\), so:\[ c^2 = 2 - 1 = 1 \quad \Rightarrow \quad c = \sqrt{1} = 1 \]The foci are \((1 \pm 1, 1) = (0, 1)\) and \((2, 1)\).

Step 7: Determine Asymptotes

For ellipses, there are no asymptotes. However, if checking for something appearing like asymptotes (considered in hyperbolas), this exercise showcases typical ellipse features without asymptotes.

Key Concepts

Ellipse Equation

An **ellipse equation** is a crucial part of understanding the geometric shape known as an ellipse, which looks like an elongated circle. Standard form equations help us identify ellipses' properties more easily. In its standard form, an ellipse's equation looks like:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here:

• \((h, k)\) represents the center of the ellipse.
• \(a^2\) and \(b^2\) are the denominators which relate to the lengths of the major and minor axes, respectively.

If \(a > b\), the ellipse is elongated along the x-axis, and if \(b > a\), it's elongated along the y-axis. In our exercise, after completing the square, we transformed the given equation to this form: \[ \frac{(x-1)^2}{2} + \frac{(y-1)^2}{1} = 1 \], indicating an ellipse centered at (1, 1) with its major axis along the x-axis.

Completing the Square

**Completing the square** is a method used to convert a quadratic equation into a perfect square trinomial. This technique is essential when rewriting conic section equations into their standard forms. To complete the square:

• Take half of the coefficient of the linear term, square it.
• Add and subtract this squared value to transform the quadratic term into a perfect square trinomial.

In our problem:

• For \(x\)-terms: \(x^2 - 2x\) became \((x-1)^2 - 1\).
• For \(y\)-terms: \(2(y^2 - 2y)\) became \(2((y-1)^2 - 1)\).

Completing the square enabled us to rewrite the equation as \(\frac{(x-1)^2}{2} + \frac{(y-1)^2}{1} = 1\), making it easier to identify the ellipse's properties.

Vertices of Ellipses

The **vertices of ellipses** are the endpoints of the longest diameter, known as the major axis. Vertices play a key role in defining the size and shape of the ellipse. To find them, you need the center \((h, k)\) and the length of the semi-major axis \(a\):

• If the major axis is horizontal, the vertices are at \((h \pm a, k)\).
• If the major axis is vertical, the vertices are at \((h, k \pm a)\).

In our specific exercise, with a center at (1, 1) and major axis along the x-axis, \(a = \sqrt{2}\). Therefore, the vertices are located at \((1 \pm \sqrt{2}, 1)\). This means our vertices are \((1+\sqrt{2}, 1)\) and \((1-\sqrt{2}, 1)\).

Foci of Ellipses

The **foci of ellipses** are two points located along the major axis, equidistant from the center, and within the ellipse. The distance between the center and each focus is crucial as it defines the ellipse's shape. To find the foci:

• Use the formula \(c = \sqrt{a^2 - b^2}\).

For our exercise, where \(a^2 = 2\) and \(b^2 = 1\), we find \(c\) by:\[ c^2 = a^2 - b^2 = 2 - 1 = 1 \]\(c = \sqrt{1} = 1\).The foci are along the major axis, horizontally from the center \((1, 1)\). Thus, the foci are at \((1+1, 1) = (2, 1)\) and \((1-1, 1) = (0, 1)\). These foci help maintain the property that the sum of the distances from any point on the ellipse to the foci is constant.