Problem 51 - College Algebra

Question

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$9 x^{2}+25 y^{2}-36 x+50 y-164=0$$

Step-by-Step Solution

Answer

Standard form of the ellipse: \((x-2)^{2}/4+(y+1)^{2}/21=1\). The foci of the ellipse are at \((2, -1 \pm \sqrt{17})\).

Step 1: Completing the Square

First, rearrange the equation to have the squares of \(x\) and \(y\) on one side: \(9(x^{2}-4x)+25(y^{2}+2y)=-164.\) Now complete the square for both \(x\) and \(y\): \(9[(x-2)^{2}-4]+25[(y+1)^{2}-1]= -164\). Simplify to find the equation in standard form: \((x-2)^{2}/4+(y+1)^{2}/21=1\). This is the standard form of an ellipse where \(h=2\), \(k=-1\), \(a^{2}=4\), and \(b^{2}=21\).

Step 2: Find the Foci

The foci of the ellipse are located at \((h, k \pm c)\) or \((h \pm c, k)\), where \(c=\sqrt{a^{2}-b^{2}}\). Since \(a^{2}

Step 3: Graph the Ellipse

The center of the ellipse is at the point \( (h,k)=(2,-1) \).The semi-major axis is \(\sqrt{21}\) units long and the semi-minor axis is \(2\) units long. The foci are at points \((2, -1 \pm \sqrt{17})\). The major axis is vertical because \(b>a\). The vertices are a distance of \(\sqrt{21}\) units above and below the center. Plot these locations to draw the graph of the ellipse.

Key Concepts

Completing the Square

To successfully complete the square, you first need to group all terms related to the same variable. For the given equation, we focus on arranging the terms around each variable: \(x\) and \(y\). We start with rearranging:

• Get terms involving \(x\) together.
• Get terms involving \(y\) together.

Here's the form before completing the square: - \(9(x^2 - 4x) + 25(y^2 + 2y) = -164\). Next, to complete the square for \(x\) and \(y\):- For \(x\), we have \(9[(x^2 - 4x + 4) - 4]\) which becomes \(9[(x-2)^2 - 4]\).- For \(y\), we have \(25[(y^2 + 2y + 1) - 1]\) which becomes \(25[(y+1)^2 - 1]\).By simplifying these, we can derive an equivalent form: \[ 9[(x-2)^2 - 4] + 25[(y+1)^2 - 1] = -164 \]. This equation lets us focus directly on isolating each square which is critical for converting to the standard form of an ellipse.

Standard Form of Ellipse

The standard form of an ellipse equation is crucial for understanding its properties. The form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) represents an ellipse centered at \((h, k)\), with the lengths of the semi-major and semi-minor axes determined by \(a\) and \(b\). From the completed square form, we simplify to find the equation in standard form: \[ \frac{(x-2)^2}{4} + \frac{(y+1)^2}{21} = 1 \]. Here, the center of the ellipse \((h, k)\) is \((2, -1)\), "a" for the semi-minor axis is \(a = \sqrt{4} = 2\), while "b" for the semi-major axis is \(b = \sqrt{21}\). Understanding this standardized form allows easy determination of the ellipse's shape and orientation. Note that because \(b > a\), the major axis is vertical.

Foci of Ellipse

The foci are special points on the ellipse that help determine its shape and dimensions. They are located along the major axis, either vertically or horizontally. To find them, we use the formula \(c^2 = b^2 - a^2\). Given that \(b^2 = 21\) and \(a^2 = 4\), solving for \(c\) gives:\[ c = \sqrt{21 - 4} = \sqrt{17} \].For an ellipse oriented as this one, with a vertical major axis:- The foci are positioned vertically from the center.- Therefore, these points are at \((2, -1 + \sqrt{17})\) and \((2, -1 - \sqrt{17})\).Recognizing the location of the foci allows for better understanding of how light or signals might behave within its reflective properties, as they naturally converge at these points.

Graphing Ellipses

Graphing an ellipse involves plotting key points like the center, the foci, and the vertices. Here, we start by identifying the center of the ellipse: - Center \((h, k) = (2, -1)\).Next, determine the lengths of axes:

• Semi-major axis, along the vertical (due to \(b > a\)), is \(\sqrt{21}\) units.
• Semi-minor axis, along the horizontal, is \(2\) units.

Locate the vertices by moving outward from the center along the axes:- Vertices occur \(\sqrt{21}\) units up and down from the center.Finally, mark the foci from earlier, located at \((2, -1 \pm \sqrt{17})\). With these points clearly plotted, sketch the ellipse remembering that it is elongated along the vertical axis.Drawing an ellipse with these characteristics communicates the shape and orientation at a glance.

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Key Concepts

Completing the Square

To successfully complete the square, you first need to group all terms related to the same variable. For the given equation, we focus on arranging the terms around each variable: \(x\) and \(y\). We start with rearranging:

• Get terms involving \(x\) together.
• Get terms involving \(y\) together.

Here's the form before completing the square: - \(9(x^2 - 4x) + 25(y^2 + 2y) = -164\). Next, to complete the square for \(x\) and \(y\):- For \(x\), we have \(9[(x^2 - 4x + 4) - 4]\) which becomes \(9[(x-2)^2 - 4]\).- For \(y\), we have \(25[(y^2 + 2y + 1) - 1]\) which becomes \(25[(y+1)^2 - 1]\).By simplifying these, we can derive an equivalent form: \[ 9[(x-2)^2 - 4] + 25[(y+1)^2 - 1] = -164 \]. This equation lets us focus directly on isolating each square which is critical for converting to the standard form of an ellipse.

Standard Form of Ellipse

The standard form of an ellipse equation is crucial for understanding its properties. The form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) represents an ellipse centered at \((h, k)\), with the lengths of the semi-major and semi-minor axes determined by \(a\) and \(b\). From the completed square form, we simplify to find the equation in standard form: \[ \frac{(x-2)^2}{4} + \frac{(y+1)^2}{21} = 1 \]. Here, the center of the ellipse \((h, k)\) is \((2, -1)\), "a" for the semi-minor axis is \(a = \sqrt{4} = 2\), while "b" for the semi-major axis is \(b = \sqrt{21}\). Understanding this standardized form allows easy determination of the ellipse's shape and orientation. Note that because \(b > a\), the major axis is vertical.

Foci of Ellipse

The foci are special points on the ellipse that help determine its shape and dimensions. They are located along the major axis, either vertically or horizontally. To find them, we use the formula \(c^2 = b^2 - a^2\). Given that \(b^2 = 21\) and \(a^2 = 4\), solving for \(c\) gives:\[ c = \sqrt{21 - 4} = \sqrt{17} \].For an ellipse oriented as this one, with a vertical major axis:- The foci are positioned vertically from the center.- Therefore, these points are at \((2, -1 + \sqrt{17})\) and \((2, -1 - \sqrt{17})\).Recognizing the location of the foci allows for better understanding of how light or signals might behave within its reflective properties, as they naturally converge at these points.

Graphing Ellipses

Graphing an ellipse involves plotting key points like the center, the foci, and the vertices. Here, we start by identifying the center of the ellipse: - Center \((h, k) = (2, -1)\).Next, determine the lengths of axes:

• Semi-major axis, along the vertical (due to \(b > a\)), is \(\sqrt{21}\) units.
• Semi-minor axis, along the horizontal, is \(2\) units.

Locate the vertices by moving outward from the center along the axes:- Vertices occur \(\sqrt{21}\) units up and down from the center.Finally, mark the foci from earlier, located at \((2, -1 \pm \sqrt{17})\). With these points clearly plotted, sketch the ellipse remembering that it is elongated along the vertical axis.Drawing an ellipse with these characteristics communicates the shape and orientation at a glance.