Problem 7 - Calculus Early Transcendentals

Question

What is the equation of the standard ellipse with vertices at \((\pm a, 0)\) and foci at \((\pm c, 0) ?\)

Step-by-Step Solution

Answer

Question: Find the equation of a standard ellipse with vertices at (±a, 0) and foci at (±c, 0). Answer: The equation of a standard ellipse with vertices at (±a, 0) and foci at (±c, 0) is given by: \(\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1\)

Step 1: Recall the general equation of a standard ellipse

The equation of a standard ellipse with center at the origin and major axis aligned along the x-axis is given by: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) Here, a is the distance from the center of the ellipse to a vertex along the major axis, and b is the distance from the center of the ellipse to a co-vertex along the minor axis. Since the vertices are given as \((\pm a, 0)\), we know the major axis is aligned along the x-axis, and the center of the ellipse will be at the origin (0,0).

Step 2: Recall the relationship between the foci, vertices, and co-vertices

For a standard ellipse, the distance between the foci and the center is related to a and b by the equation: \(c^2 = a^2 - b^2\) In this problem, we are given that the foci are at the points \((\pm c,0)\).

Step 3: Find the value of b

Since we are given values for a and c, we can use the relationship between the foci, vertices, and co-vertices to find the value of b. From step 2, we know that: \(c^2 = a^2 - b^2\) Rearrange the equation to solve for b^2: \(b^2 = a^2 - c^2\)

Step 4: Substitute the values of a and c into the general equation

Now that we have an expression for b^2 in terms of a and c, we can substitute these values into the general ellipse equation obtained in Step 1: \(\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1\) This is the equation of the standard ellipse with vertices at \((\pm a, 0)\) and foci at \((\pm c, 0)\).