Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$ 4 x^{2}+9 y^{2}-32 x-36 y+64=0 $$

Step-by-Step Solution

Answer

Vertices are (7, 2) and (1, 2); foci are approximately (4±√5, 2).

Step 1: Rearrange the Equation

Start by rearranging the given equation to group the same variables together:\[ 4x^2 - 32x + 9y^2 - 36y + 64 = 0 \]

Step 2: Complete the Square

For the equation to take the standard form of an ellipse, complete the square for both the \(x\) and \(y\) terms.Start with the \(x\) terms:Factor out the coefficient of \(x^2\):\[ 4(x^2 - 8x) \]Complete the square:\[ 4(x^2 - 8x + 16 - 16) = 4((x-4)^2 - 16) \]Next, the \(y\) terms:Factor out the coefficient of \(y^2\):\[ 9(y^2 - 4y) \]Complete the square:\[ 9(y^2 - 4y + 4 - 4) = 9((y-2)^2 - 4) \]

Step 3: Simplify the Equation

Substitute the completed squares back into the equation:\[ 4((x-4)^2 - 16) + 9((y-2)^2 - 4) + 64 = 0 \]Simplify and organize:\[ 4(x-4)^2 - 64 + 9(y-2)^2 - 36 + 64 = 0 \]\[ 4(x-4)^2 + 9(y-2)^2 = 36 \]

Step 4: Divide by 36 to Standardize

Divide the entire equation by 36 to express it in the standard form of an ellipse:\[ \frac{4(x-4)^2}{36} + \frac{9(y-2)^2}{36} = 1 \]This simplifies to:\[ \frac{(x-4)^2}{9} + \frac{(y-2)^2}{4} = 1 \]

Step 5: Identify the Center and Axes

From the standard form \( \frac{(x-4)^2}{9} + \frac{(y-2)^2}{4} = 1 \), we identify:- Center: \((h, k) = (4, 2)\)- Lengths of the semi-major and semi-minor axes: \(a = 3\), \(b = 2\)- Since \(a > b\), the ellipse is wider along the \(x\)-axis.

Step 6: Calculate the Vertices

The vertices along the major axis \((a = 3)\) are:- \((h+a, k) = (4+3, 2) = (7, 2)\)- \((h-a, k) = (4-3, 2) = (1, 2)\)

Step 7: Calculate the Foci

The foci are located \(c\) units from the center,where \(c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5}\).The foci along the major axis are:- \((h+c, k) = (4+\sqrt{5}, 2)\)- \((h-c, k) = (4-\sqrt{5}, 2)\)

Step 8: Sketch the Graph

Draw an ellipse centered at \((4, 2)\) with vertices at \((7, 2)\) and \((1, 2)\), and indicated foci at approximately \((4+\sqrt{5}, 2)\) and \((4-\sqrt{5}, 2)\). Remember that it is wider along the \(x\)-axis since \(a > b\).

Key Concepts

Vertices

Finding the vertices of an ellipse is a crucial step when working with its geometry. Vertices are essentially the points where the ellipse intersects the axes of symmetry.
For ellipses oriented horizontally, like the one we have in our equation \[\frac{(x-4)^2}{9} + \frac{(y-2)^2}{4} = 1\], the major axis is along the horizontal direction.

• The center of this ellipse is at \((h, k) = (4, 2)\).
• Since the major axis length is represented by \(2a\), here \(a = 3\).
• The vertices are therefore found at \((h+a, k)\) and \((h-a, k)\), leading to \((7, 2)\) and \((1, 2)\).

These vertices give the widest points of the ellipse, and for a horizontal ellipse, they will be to the left and right of the center.

Foci

Just as vertices mark significant points on an ellipse, the foci are pivotal for its definition as well. An ellipse can be considered the set of all points where the sum of distances to two fixed points, called foci, remains constant.
For our ellipse, \(c\) represents the distance of each focus from the center, calculated using \(c = \sqrt{a^2 - b^2}\).
In our case, since \(a = 3\) and \(b = 2\),

• \(c = \sqrt{9 - 4} = \sqrt{5}\).
• The foci are positioned along the major axis, \((h+c, k)\) and \((h-c, k)\), resulting in approximately \((4+\sqrt{5}, 2)\) and \((4-\sqrt{5}, 2)\).

Understanding the foci helps in visualizing how narrow or stretched an ellipse appears. The closer the foci are to the center, the more circle-like the ellipse looks.

Completing the Square

To convert an algebraic equation into the more usable form of an ellipse, completing the square is a vital algebraic technique. It involves rewriting quadratic terms to reveal the structure needed for the standard form.
Starting with the equation:\[4x^2 - 32x + 9y^2 - 36y + 64 = 0\],
we complete the square separately for the \(x\) and \(y\) terms:

• For \(x^2\): Factor out the 4: \[4(x^2 - 8x)\] Complete the square, converting into \[4((x-4)^2 - 16)\].
• For \(y^2\): Factor out the 9: \[9(y^2 - 4y)\] Complete the square, converting into \[9((y-2)^2 - 4)\].

This step is fundamental as it transforms a confusing quadratic equation into an ellipse recognizable by its geometric properties.

Standard Form of Ellipse

The standard form of an ellipse is key to understanding and identifying its geometric properties effectively. For an ellipse centered at \((h, k)\), the standard form is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.\]In our given problem, the equation:
\[\frac{(x-4)^2}{9} + \frac{(y-2)^2}{4} = 1\]
shows an ellipse centered at \((4, 2)\).

• The term \((x-4)^2/9\) indicates a semi-major axis length of \(a = 3\).
• The term \((y-2)^2/4\) indicates a semi-minor axis length of \(b = 2\).

The standard form clearly presents the essential attributes such as center, axes lengths, and orientation, enabling us to draw the ellipse accurately. It's what makes the ellipse analyzable and useful in mathematics.