Problem 45 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ 4(x+1)^{2}+9(y+1)^{2}=36 $$

Step-by-Step Solution

Answer

The standard form is \(\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\), an ellipse with center \((-1,-1)\), major axis 3 in \(x\) direction, minor axis 2 in \(y\) direction.

Step 1: Identify the Equation Type

The given equation is \(4(x+1)^2 + 9(y+1)^2 = 36\). This equation resembles the general form of an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). This means the graph is an ellipse since there are squared terms for both \(x\) and \(y\) with different coefficients.

Step 2: Convert to Standard Form

Start by dividing the entire equation by 36 to match the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). This gives:\[\frac{4(x+1)^2}{36} + \frac{9(y+1)^2}{36} = 1\]Simplifying the fractions, we get:\[\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\]

Step 3: Identify Center, Axes, and Radii

From the standard form \(\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\), identify the center as \((-1, -1)\). The denominator under \((x+1)^2\) is 9, so the semi-major axis in the \(x\) direction has length 3 (since \(\sqrt{9} = 3\)). The denominator under \((y+1)^2\) is 4, so the semi-minor axis in the \(y\) direction has length 2 (since \(\sqrt{4} = 2\)).

Step 4: Graph the Ellipse

To graph the ellipse, first plot the center at \((-1,-1)\). From the center, move 3 units to the left and right along the \(x\)-axis to indicate the ends of the major axis, and 2 units up and down along the \(y\)-axis to indicate the ends of the minor axis. Draw an ellipse passing through these points, with the longer axis along the horizontal \(x\)-axis.

Key Concepts

Standard Form Equation

When working with ellipses, having the equation in its standard form is crucial. The standard form of an ellipse's equation is given by:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here, \(h, k\) is the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively.
To convert an equation, such as \(4(x+1)^2 + 9(y+1)^2 = 36\), to standard form, we need to get a 1 on the right-hand side. This is achieved by dividing every term by 36, the constant on the right. Doing so transforms the equation into:\[\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\]
This process allows us to clearly identify the components needed to graph the ellipse accurately.

Center of Ellipse

The center of an ellipse is a critical concept to understand because it helps to orient the graph correctly. From the standard form equation:\[\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\]we can see that the center of the ellipse is at \((-1, -1)\).
In general, the center is represented by the \(h\) and \(k\) in the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
The expression \(x+1\) indicates a shift of the center to the left by one unit from the origin, and \(y+1\) indicates a downward shift by one.
This central point acts as the reference for the placement of the semi-major and semi-minor axes and overall layout of the ellipse on the coordinate plane.

Semi-Major Axis

In an ellipse, the semi-major axis represents the longest radius. It extends from the center to the farthest point on the ellipse.In our standard form equation, \(\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\), the larger of the two denominators (9 in this case) sits under the \(x\)-term.
This arrangement signals that the semi-major axis is parallel to the \(x\)-axis.
To find its length, take the square root of 9, giving us 3. So, the ellipse stretches 3 units in each direction from the center along the \(x\)-axis.
This means the total length of the major axis is 6 units, emphasizing the horizontal stretch of this particular ellipse.

Semi-Minor Axis

Slightly shorter than the semi-major axis, the semi-minor axis is the ellipse's shortest radius, extending from the center.In the equation \(\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1\), the smaller denominator is 4, which is found under the \(y\)-term.
This implies that the semi-minor axis is oriented along the \(y\)-axis.
Its length is determined by taking the square root of 4, resulting in a length of 2.
This 2-unit measurement from the center in both upward and downward directions along the \(y\)-axis defines the compact vertical stretch of the ellipse.
Hence, understanding the semi-minor axis is key to drawing the ellipse accurately.