Question

Sketch the graph of the following ellipses. Plot and label the coordinates of the vertices and foci, and find the lengths of the major and minor axes. Use a graphing utility to check your work. $$12 x^{2}+5 y^{2}=60$$

Short answer

Question: Sketch the graph of the ellipse given by the equation 12x^2 + 5y^2 = 60. Answer: To sketch the graph of this ellipse, first find the coordinates of the vertices and the foci. The vertices are at points (±√5, 0) and (0, ±2√3), and the foci are at points (0, ±√7). Plot these points, and use them to draw an ellipse centered at the origin, with major and minor axes of 2√3 and √5, respectively.

Step-by-step solution

Find a and b

We need to rewrite the given equation in the standard form of ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ To do this, divide both sides of the equation by 60: $$\frac{x^2}{5}+\frac{y^2}{12}=1$$ Now, we can see that \(a^2=5\) and \(b^2=12\). This means that \(a=\sqrt{5}\) and \(b=\sqrt{12}=2\sqrt{3}\).

Find the Coordinates of the Vertices

For the ellipse in this form, the vertices lie at the points (±a, 0) and (0, ±b). Therefore, the coordinates of the vertices are: $$(\pm\sqrt{5}, 0) \text{ and } (0, \pm2\sqrt{3})$$

Calculate the length of the Semi-major and Semi-minor Axis

Since the denominator under \(y^2\) is larger, this ellipse has a vertical major axis, and b is the length of the semi-major axis. Therefore, the length of the semi-major axis is \(2\sqrt{3}\), and the length of the semi-minor axis is \(\sqrt{5}\).

Find the Coordinates of the Foci

Since the major axis is vertical, the foci have coordinates (0, ±c), where \(c = \sqrt{a^2-b^2}\). We calculate c as: $$c=\sqrt{12-5}=\sqrt{7}$$ Thus, the foci have coordinates: $$(0, \pm\sqrt{7})$$

Sketch the Graph and Check with a Graphing Utility

Now, with all the necessary information, we can plot the coordinates of the vertices and foci, draw an ellipse centered at the origin, and plot the major and minor axes. The coordinates of the vertices are: $$(\pm\sqrt{5}, 0) \text{ and } (0, \pm2\sqrt{3})$$ The coordinates of the foci are: $$(0, \pm\sqrt{7})$$ To check our work, use a graphing utility (like Geogebra or Desmos) to plot 12x^2+5y^2=60. You should see a vertically elongated ellipse centered at the origin, with vertices and foci as calculated above.

Key concepts

Vertices of an Ellipse

In the world of ellipses, the vertices are very important as they determine the overall shape. The vertices of an ellipse are the farthest points on the major and minor axes. In simpler terms, they are the points where the ellipse stretches the furthest both horizontally and vertically. For an ellipse centered at the origin, the equation can tell us the exact location of these vertices.
Typically, an equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). If the larger denominator is under \( x^2 \), it means the ellipse stretches farther horizontally (horizontally oriented), and its vertices will be at \( (\pm a, 0) \). Conversely, if the larger denominator is under \( y^2 \), as it is in our original problem, the ellipse is vertically oriented, and the vertices are found at \( (0, \pm b) \).
In our problem, \( b^2 = 12 \), so \( b = 2\sqrt{3} \). This means our vertices are at \( (0, \pm 2\sqrt{3}) \), stretched along the y-axis.

Graphing Ellipses

Graphing ellipses can be a fun way to visually understand the geometric properties of these shapes. The first step in graphing an ellipse is to convert its equation into standard form, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). If an equation like \( 12x^2 + 5y^2 = 60 \) doesn't seem to fit, we simplify it by dividing through by the constant to match this form.
Once in standard form, calculate \( a^2 \) and \( b^2 \) to find \( a \) and \( b \), which define how far the ellipse extends along the x and y axes. For our problem, we're dealing with an equation that allows us to identify the respective lengths of the semi-major and semi-minor axes.
Plot the vertices and draw curves smoothly connecting them to form an elongated oval. For verification, tools like Desmos or Geogebra can help visualize and check our plotted ellipse against the equation.

Foci of an Ellipse

The foci are two special points inside an ellipse. They help define its width and shape. For any point on an ellipse, the sum of the distances to each focus is constant. This property is fundamental and can be used practically to find these points.
In our ellipse, with the orientation given by the larger \( y \) term, we calculate the distance of the foci using the formula \( c = \sqrt{b^2 - a^2} \). Here \( b^2 = 12 \) and \( a^2 = 5 \), so \( c = \sqrt{12 - 5} = \sqrt{7} \). This gives us the foci located at \( (0, \pm \sqrt{7}) \).
The foci are vital in plotting the ellipse accurately, ensuring the correct elongation based on their position.

Axes of an Ellipse

Understanding the axes of an ellipse helps fully grasp its dimension and orientation. An ellipse has two key axes: the major axis and the minor axis. The major axis is the longer axis, while the minor is the shorter one. Their measurements are derived from the terms \( a \) and \( b \), which are squared in the ellipse equation.
For the given problem, the vertical major axis indicates that the ellipse is taller than it is wide. We calculate the length of the major axis as \( 2b = 2 \times 2\sqrt{3} = 4\sqrt{3} \), while the length of the minor axis is \( 2a = 2 \times \sqrt{5} \).
These calculations give us a complete view of the ellipse's proportions, helping us sketch it precisely on graph paper by marking these dimensions.