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Chapter 10: Problem 26

Find the standard form of the equation of the ellipse with the given characteristics. Foci: (0,0),(4,0)\(;\) major axis of length 6

Short Answer

Expert verified
The standard form of the equation of the ellipse is \(\frac{(x - 2)^2}{9} + \frac{y^2}{5} = 1\)

Step by step solution

01

Determine the Center of the Ellipse

The center of the ellipse is the midpoint of the line segment between the two foci. As foci are given as (0,0) and (4,0), the center of the ellipse will be the average of the foci coordinates, i.e. \(\frac{0+4}{2} , \frac{0+0}{2}\) which gives (2, 0).
02

Determine a

The length of the major axis is 6 units. This distance is equal to 2\(a\), thus to find \(a\) we need to divide the length of the major axis by 2, which is \(\frac{6}{2}\) = 3.
03

Determine the Value of b

The distance from the center to each focus is \(\sqrt{a^2 - b^2}\). Our foci are at distances 2 units to the right and left of the center, so \(\sqrt{a^2 - b^2}\) = 2. Substitute \(a = 3\) into the equation, thus we have \(\sqrt{9 - b^2}\) = 2. Square both sides, we get \(9 - b^2 = 4\), so \(b^2\) = 5. Since \(b\) is always positive, its value is \(\sqrt{5}\).
04

Write the Equation of the Ellipse

Substitute the values of the center, \(a\) and \(b\) into the standard form of the equation of the ellipse. Hence, we get \(\frac{(x - 2)^2}{9} + \frac{y^2}{5} = 1\).

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Most popular questions from this chapter

Find the exact value of the trigonometric expression when \(u\) and \(v\) are in Quadrant IV and \(\sin u=-\frac{3}{5}\) and \(\cos v=1 / \sqrt{2}\). $$\sin (u-v)$$

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