Problem 15 - Calculus Early Transcendentals

Question

Find the vertices and foci of the ellipse and sketch its graph. $$9 x^{2}-18 x+4 y^{2}=27$$

Step-by-Step Solution

Answer

Vertices: (1, 3), (1, -3), (3, 0), (-1, 0); Foci: (1, √5), (1, -√5).

Step 1: Write the Standard Form of the Ellipse Equation

The given equation is \(9x^2 - 18x + 4y^2 = 27\). First, rewrite this equation in the standard form of an ellipse equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Complete the square for the \(x\) terms and then simplify.

Step 2: Complete the Square for the x-terms

Begin by factoring the \(x\) terms: \(9(x^2 - 2x) + 4y^2 = 27\). Complete the square inside the parenthesis: \(x^2 - 2x = (x-1)^2 - 1\). Substitute back into the equation: \(9((x-1)^2 - 1) + 4y^2 = 27\). Simplify to get: \(9(x-1)^2 - 9 + 4y^2 = 27\). Add 9 on both sides to balance: \(9(x-1)^2 + 4y^2 = 36\).

Step 3: Divide by 36 to Achieve Standard Form

Rewrite the equation dividing every term by 36 for the standard form: \(\frac{9(x-1)^2}{36} + \frac{4y^2}{36} = \frac{36}{36}\). Simplify to: \(\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1\). This represents a vertical ellipse centered at \((h, k) = (1, 0)\).

Step 4: Identify Vertices and Semi-major/Minor Axes

For \(\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1\), the semi-major axis is vertical with \(b^2 = 9\) \((b = 3)\), and the semi-minor axis is horizontal with \(a^2 = 4\) \((a = 2)\). The vertices are at \((1, 3)\), \((1, -3)\), \( (1 + 2, 0)\) and \((1 - 2, 0)\).

Step 5: Calculate Foci

For the ellipse, find \(c\) where \(c^2 = b^2 - a^2\): \(c^2 = 9 - 4 = 5\). Hence, \(c = \sqrt{5}\). The foci are located \(c\) units from the center \((1, 0)\) along the y-axis, at \((1, \sqrt{5})\) and \((1, -\sqrt{5})\).

Step 6: Sketch the Ellipse

You now have all the key components: center \((1, 0)\), vertices \((1, 3)\) and \((1, -3)\), co-vertices \((3, 0)\) and \((-1, 0)\), and foci \((1, \sqrt{5})\) and \((1, -\sqrt{5})\). Use these points to sketch a centered ellipse with horizontal length \(4\) and vertical length \(6\).

Key Concepts

Vertices

In an ellipse, vertices are pivotal points that define the extent of its longest axis, also known as the major axis. For the given problem, we first express the ellipse equation in its standard form, which helps to identify the vertices. The vertices of an ellipse lie along the major axis, centered around the ellipse's midpoint or center.

In the context of the equation \[\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1,\]we recognize that the major axis is vertical because the larger denominator (9) is underneath the \(y\)-variable. This indicates that the distance from the center to the vertices along the \(y\)-axis is \(b\) units, where \(b = 3.\) This gives us the vertices at \((1, 3)\) and \((1, -3)\).

Knowing these points is crucial for sketching the ellipse since they indicate the ellipse's maximum stretch along the vertical direction.

Foci

Ellipses are not just defined by their oval shape and vertices. Another important characteristic of an ellipse is its foci (singular: focus). These are points inside the ellipse that are crucial in determining its geometric properties.

For the given ellipse \[\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1,\]we use the relationship \(c^2 = b^2 - a^2\) to determine the distance of the foci from the center. Here, \(b^2 = 9\) and \(a^2 = 4\), giving us:\[c^2 = 9 - 4 = 5,\]and thus \(c = \sqrt{5}.\)

The foci are located \(c\) units from the center along the major axis direction, which means they are to be found at \((1, \sqrt{5})\) and \((1, -\sqrt{5})\). Understanding the location of the foci is essential for understanding how the ellipse retains its shape.

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This is crucial in converting the general form of a conic section (like an ellipse) into its standard form.

In the given exercise, the equation \[9x^2 - 18x + 4y^2 = 27\]requires us to complete the square for the \(x\) terms. We first factor out the coefficient of \(x^2\), which is 9:\[9(x^2 - 2x) + 4y^2 = 27.\]

Completing the square for \(x^2 - 2x\), we have:\[x^2 - 2x = (x-1)^2 - 1.\] Substituting back, the equation becomes:\[9((x-1)^2 - 1) + 4y^2 = 27.\]This simplifies further to \[9(x-1)^2 + 4y^2 = 36,\]laying groundwork for rewriting in standard form.

By completing the square, we can more easily identify properties like the ellipse's center, axes lengths, and orientation.

Standard Form of Ellipse

The standard form of an ellipse equation provides a clear and concise representation of its geometric properties. This form is expressed as:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where \((h, k)\) is the center of the ellipse, \(a\) is the semimajor axis length, and \(b\) is the semiminor axis length.

From the general quadratic equation \[9x^2 - 18x + 4y^2 = 27,\]we transformed this into its standard form through completing the square:\[\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1.\] This clearly shows that the center of the ellipse is at \((1, 0)\), with \(a = 2\) and \(b = 3\). The major axis is along the \(y\)-axis, hence vertical, due to \(b > a\).

The standard form allows us to easily read off important features about the ellipse which are critical for graphing it correctly.

Ellipse Equation

Understanding the ellipse equation is the foundational step in solving problems involving ellipses. An equation involving squared terms represents an ellipse if it can be transformed to fit the standard form mentioned earlier.

For the exercise equation:\[9x^2 - 18x + 4y^2 = 27,\]we started by rearranging and simplifying it into its standard form. The process of transforming it involved completing the square, which allowed us to isolate squared terms. Ultimately, this provided us with the ellipse equation:\[\frac{(x-1)^2}{4} + \frac{y^2}{9} = 1.\]

Accurately identifying and rewriting the proper form of the ellipse equation is crucial, not just for computations, but also for graphing. Only then can characteristics such as vertices, foci, and axes be determined. Grasping the basics of forming and understanding these equations will greatly aid in mastering more complex ellipse-related problems in mathematics.