Problem 29 - College Algebra

Question

Find the standard form of the equation of each ellipse satisfying the given conditions. Foci: \((-2,0),(2,0) ; y\) -intercepts: \(-3\) and 3

Step-by-Step Solution

Answer

The standard form of the equation of the ellipse is \(\frac{x^2}{40} + \frac{y^2}{36} = 1\)

Step 1: Identify the center, major axis and minor axis

The center of ellipse is the origin (0,0) because the foci are symmetric about the y-axis. The length of the minor axis is the absolute value of the y-intercepts \(-3-3=6\). The distance between the foci is \(2-(-2)=4\), which represents \(2c\). Thus, \(c=2\). The length of the major axis (2a) can then be found using the relationship \(a^2=b^2+c^2\).

Step 2: Calculate the length of the major axis

Using the equation \(a^2=b^2+c^2\), substitute the value of \(b=2\) and \(c=2\) to determine the value of \(a\). So \(a^2=6^2+2^2=40\). Therefore, \(a=\sqrt{40}=2\sqrt{10}\)

Step 3: Write the equation of the ellipse

The standard form for the equation of an ellipse centered at the origin with major axis along the x-axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Substitute \(a=2\sqrt{10}\) and \(b=6\) into the equation to get \(\frac{x^2}{(2\sqrt{10})^2} + \frac{y^2}{6^2} = 1\), which simplifies to \(\frac{x^2}{40} + \frac{y^2}{36} = 1\)