Problem 22 - Thomas Calculus

Question

Exercises \(17-24\) give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch. $$ 9 x^{2}+10 y^{2}=90 $$

Step-by-Step Solution

Answer

The ellipse's standard form is \(\frac{x^2}{10} + \frac{y^2}{9} = 1\), with foci at \((1, 0)\) and \((-1, 0)\).

Step 1: Divide to Simplify the Equation

The given equation of the ellipse is \(9x^2 + 10y^2 = 90\). To begin, divide every term by \(90\) to simplify the equation.\[\frac{9x^2}{90} + \frac{10y^2}{90} = \frac{90}{90}\]Simplifying this gives:\[\frac{x^2}{10} + \frac{y^2}{9} = 1\]

Step 2: Identify the Ellipse Parameters

The equation \(\frac{x^2}{10} + \frac{y^2}{9} = 1\) is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a^2 = 10\) and \(b^2 = 9\), with \(a = \sqrt{10}\) and \(b = 3\).Since \(a^2 > b^2\), the major axis is along the \(x\)-axis.

Step 3: Find the Foci of the Ellipse

The foci of an ellipse are determined by the formula \(c = \sqrt{a^2 - b^2}\).Using our values, \(c = \sqrt{10 - 9} = \sqrt{1} = 1\).Thus, the foci are located at \((\pm c, 0)\), or \((\pm1, 0)\).

Step 4: Sketch the Ellipse

To sketch the ellipse, draw the major axis with length \(2a = 2\sqrt{10}\) along the \(x\)-axis, and the minor axis with length \(2b = 6\) along the \(y\)-axis.Plot and label the foci at \((1, 0)\) and \((-1, 0)\).Draw the ellipse around the origin ensuring it is centered at (0,0) as per the standard form.

Key Concepts

Ellipse Sketching

Sketching an ellipse is an important skill that allows you to visualize the properties of the ellipse from its equation. Once you have transformed the original equation into its standard form, the next step is to draw the ellipse.

Here’s how you do it:

• Identify the lengths of the major and minor axes. The major axis is the longest diameter, running through the center of the ellipse along the x or y-axis, depending on which squared term has the larger denominator.
• For this exercise, with equation \(\frac{x^2}{10} + \frac{y^2}{9} = 1\), the major axis is along the x-axis.
• The length of the major axis is \(2a = 2\sqrt{10}\).
• Meanwhile, the minor axis runs perpendicular to the major axis and has a length of \(2b = 6\).

To accurately sketch, draw these axes crosswise at the center of the graph and make sure the ellipse is symmetric. Adding the foci will further assist in making the sketch precise and complete.

Ellipse Foci

The foci of an ellipse are two special points that lie on the major axis. They play a crucial role in the definition and properties of an ellipse. The sum of the distances from any point on the ellipse to these two foci is constant.

To find the foci, use the equation \(c = \sqrt{a^2 - b^2}\), where \(a\) and \(b\) represent the semi-major and semi-minor axes respectively.

• In our problem, we have \(a^2 = 10\) and \(b^2 = 9\), leading to \(c = \sqrt{10-9} = 1\).
• The foci are then located at \((\pm1, 0)\) along the major axis (x-axis) because the major axis is horizontal.

These foci positions are critical in ensuring that your sketch of the ellipse is both accurate and detailed.

Standard Form of an Ellipse

The standard form of an ellipse is essential for understanding its geometric properties. By converting the given ellipse equation into standard form, you can easily determine its key characteristics.

An ellipse equation is typically written as:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
if the ellipse is oriented horizontally, and vice versa if oriented vertically.

• In the problem, the equation \(9x^2 + 10y^2 = 90\) converts into \(\frac{x^2}{10} + \frac{y^2}{9} = 1\).
• This conversion is accomplished by dividing each term by 90, simplifying the equation to match the standard form.
• With \(a^2 = 10\) and \(b^2 = 9\), it's clear that the major axis is along the x-direction.

Understanding this form is vital for analyzing and sketching the ellipse accurately, revealing its true size and orientation.