Problem 27 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the ellipse with the given characteristics. $$\text { Center: }(3,2) ; a=3 c ; \text { foci: }(1,2),(5,2)$$

Step-by-Step Solution

Answer

The standard form equation of the ellipse is: \((x - 3)^2/36 + (y - 2)^2/32 = 1\)

Step 1: Determine the value of c

Based on the given foci, the value of c can be obtained from the distance between a given foci and the center. This is achieved by subtracting the x-coordinate of a foci from the x-coordinate of the center, \(c = | 3 - 1 | = 2\), since the y-coordinates of the foci and the center are the same.

Step 2: Find the value of a

Given the relationship \(a=3c\), we substitute the value of c into this equation to get the value of a, \(a=3*2=6\).

Step 3: Calculate the value of b

We can find b by substituting the values of a and c found in steps 1 and 2 into the equation \(c^2 = a^2 - b^2\). Therefore, \(b = \sqrt{a^2 - c^2} = \sqrt{(6)^2 - (2)^2} = \sqrt{32}\), which simplifies to \(b = 4\sqrt{2}\).

Step 4: Write the standard form of the equation of the ellipse

Using the center coordinates (3,2), and the values for a and b found in the previous steps, the standard form equation of an ellipse is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). Substituting the values of h=3, k=2, a=6, and b=4sqrt(2) into the equation, we obtain the standard form equation of the ellipse as \((x - 3)^2/6^2 + (y - 2)^2/(4\sqrt{2})^2 = 1\), or in simplified form, \((x - 3)^2/36 + (y - 2)^2/32 = 1\).