Problem 47 - Intermediate Algebra

Question

Identify whether equation, when graphed, will be a parabola, circle, ellipse, or hyperbola. Sketch the graph of equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. \(\frac{(x-1)^{2}}{49}+\frac{(y+2)^{2}}{25}=1\)

Step-by-Step Solution

Answer

The equation represents an ellipse centered at \((1, -2)\).

Step 1: Recognize the form of the equation

The given equation is \( \frac{(x-1)^{2}}{49} + \frac{(y+2)^{2}}{25} = 1 \). This is in the form \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), which represents the equation of an ellipse.

Step 2: Identify the center of the ellipse

The center \((h, k)\) of an ellipse described by \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \) is given by the values \( h = 1 \) and \( k = -2 \). Therefore, the center of this ellipse is \((1, -2)\).

Step 3: Determine the lengths of the semi-major and semi-minor axes

In the equation, \( a^{2} = 49 \) and \( b^{2} = 25 \). Therefore, \( a = 7 \) and \( b = 5 \). Since \( a > b \), the semi-major axis is vertical with a length of 7 and the semi-minor axis is horizontal with a length of 5.

Step 4: Sketch the graph

To sketch the ellipse, plot the center at \((1, -2)\). From there, mark points 7 units up and down for the ends of the semi-major axis, and 5 units left and right for the ends of the semi-minor axis. Connect these points smoothly to form the ellipse shape.

Key Concepts

Ellipse

An ellipse is a type of conic section that looks like an elongated circle. It represents the set of all points where the sum of the distances from two fixed points, known as foci, is constant. The ellipse can be identified in its general equation form:

• \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)

Here, \((h, k)\) is the center, and \(a\) and \(b\) are the distances from the center to the vertices along the major and minor axes, respectively. For our equation \( \frac{(x-1)^2}{49} + \frac{(y+2)^2}{25} = 1 \), this is definitely an ellipse.
The equation tells us that the ellipse has a center at \((1, -2)\), with semi-major and semi-minor axes lengths determined by the square roots of 49 and 25. This specific configuration, with a greater vertical axis length \(a=7\) than horizontal \(b=5\), indicates an ellipse stretched more in the vertical direction.

Graphing Equations

Graphing the equation of a conic section such as an ellipse involves plotting the key points: the center and the extents of its axes. Here's how you graph the ellipse:

• Start by identifying the center of the ellipse using the coordinates \((h, k)\), which are \((1, -2)\) for our specific equation.
• Determine the lengths of the semi-major \(a = 7\) and semi-minor \(b = 5\) axes from the equations \(a^2 = 49\) and \(b^2 = 25\).

Once these details are known, you can sketch the ellipse:

• Mark the center at \((1, -2)\).
• From the center, count 7 units up and down to mark the ends of the major axis and 5 units left and right for the minor axis.
• Connect these points using a smooth, curving line to form the ellipse shape.

This systematic approach helps in accurately representing the ellipse on a graph.

Centers of Conic Sections

The center of conic sections, especially in the case of ellipses and circles, is a crucial starting point for understanding their properties and graphing them accurately. For ellipses, the center \((h, k)\) is not only pivotal for placing the shape on the graph but also for determining the orientation and alignment of its axes.
In our elliptical equation \( \frac{(x-1)^2}{49} + \frac{(y+2)^2}{25} = 1 \), the center is calculated to be \((1, -2)\) by identifying the values substituted for \(x\) and \(y\) in the "\((x-h)\)" and "\((y-k)\)" terms in the standard ellipse equation form.
Understanding the center is essential because:

• It helps establish the initial point for drafting the ellipse.
• From the center, you measure out the semi-major \(a\) and semi-minor \(b\) axes to accurately plot the ellipse.
• In symmetrical conic sections like the ellipse, the center equally divides the sections across its axes.

By identifying the center correctly, we create a stable foundation upon which to build the rest of the graphical representation.