Problem 15 - Intermediate Algebra

Question

Find \(h, k, a,\) and \(b: \frac{(x+8)^{2}}{100}+\frac{(y-6)^{2}}{144}=1\)

Step-by-Step Solution

Answer

The values are: \(h = -8\), \(k = 6\), \(a = 10\), \(b = 12\).

Step 1: Identify the form of the equation

The given equation \(\frac{(x+8)^{2}}{100}+\frac{(y-6)^{2}}{144}=1\) is in the standard form of the equation of an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \(h, k\) represent the center coordinates of the ellipse, and \(a, b\) represent the lengths of the semi-major and semi-minor axes.

Step 2: Match the equation with the standard form

Compare the given equation with the standard form. In \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), \(h\) is the number paired with \(x\) with a change in sign, \(k\) is the number paired with \(y\) with a change in sign, \(a^2\) is the denominator under the \(x\)-term, and \(b^2\) is the denominator under the \(y\)-term.

Step 3: Determine the values of \(h\) and \(k\)

From the equation \((x+8)\), we find \(h\) by changing the sign of \(+8\) to \(-8\). From \((y-6)\), we obtain \(k\) directly as \(6\). Thus, \(h = -8\) and \(k = 6\).

Step 4: Find the values of \(a\) and \(b\)

The value of \(a^2\) is \(100\), so \(a = \sqrt{100} = 10\). The value of \(b^2\) is \(144\), so \(b = \sqrt{144} = 12\). Thus, \(a = 10\) and \(b = 12\).

Step 5: Summarize the values

Now that we have determined all the values, the center of the ellipse is \((-8, 6)\), and the lengths of the semi-axes are \(a = 10\) and \(b = 12\).

Key Concepts

Standard Form of Ellipse

The standard form of an ellipse in mathematics is represented by the equation: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]. This equation helps identify the geometric properties of an ellipse, such as its center and axes lengths. Here,

• \( h \): x-coordinate of the center
• \( k \): y-coordinate of the center
• \( a^2 \): square of the length of the semi-major axis
• \( b^2 \): square of the length of the semi-minor axis

It’s important to note that the ellipse differs from circles primarily in having two different axes lengths. While a circle has a single radius, an ellipse stretches differently along its axes, creating an oval shape. Recognizing the difference in equations, such as substituting equal radii in circles for differing squared terms here, will help distinguish between these conic sections.

Ellipse Center

To find the center of an ellipse given in standard form, you look at the values of \( h \) and \( k \) in the equation. For example, in the equation \[ \frac{(x+8)^2}{100} + \frac{(y-6)^2}{144} = 1 \],

• The value "\(+8\)" in the \((x + 8)^2\) term, actually indicates \( h = -8 \)
• The "\(-6\)" in \((y - 6)^2\) indicates \( k = 6 \)

Thus, the center of this ellipse is \((-8, 6)\). The signs of \( h \) and \( k \) should always be noted through the formula \((x-h)^2\) and \((y-k)^2\), which essentially flips their apparent signs. Understanding these adjustments clarifies correct plotting for graphing an ellipse's precise center.

Semi-Major Axis

In an ellipse, the semi-major axis is essentially the longest radius, or half the longest diameter, stretching from the center to the perimeter. In the standard form equation \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \], the term that has the larger denominator today identifies the semi-major axis.
In the provided example, where \( a^2 = 100 \) and \( b^2 = 144 \), the semi-major axis corresponds to the term with \( b^2 \). Therefore:

• The semi-major axis length \( b \) is \( \sqrt{144} = 12 \)

It’s pivotal always to strip away square roots effectively, as ellipses’ eccentric nature requires each axis to be profoundly different, especially when graphed.

Semi-Minor Axis

The semi-minor axis stands as the ellipse's shorter radius, or one-half of its smallest diameter. Identifying it follows the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Whichever denominator is smaller between \( a^2 \) and \( b^2 \) aligns with the semi-minor.
In our example, with \( a^2 = 100 \), the semi-minor axis length \( a \) is derived thus:

• \( a = \sqrt{100} = 10 \)

As semi-minor axes define the ellipse's width, they’re a central consideration when visualizing complete shapes. Though shorter, these axes help encapsulate how much less steep the ellipse’s curvature is from endpoint to endpoint.