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Chapter 6: Problem 22

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 \((x - 2)^2/3^2 + y^2/5 = 1\).

Step by step solution

01

Determine the Center, Semi-Major Axis, and Focus Distance

The foci are given by (0,0) and (4,0), which are symmetric about the y-axis, therefore, the center of the ellipse will be the midpoint of the two foci, which is at (2,0). The major axis is given to be 6 units long, thus the semi-major axis \(a\) is half of this, or 3 units. A focus is at a distance \(c\) from the center, given by c = 2 units (from the coordinates of the foci).
02

Calculate the Semi-Minor Axis

The relationship between \(a\), \(b\) (the semi-minor axis), and \(c\) (the distance from the center to a focus) is given by the equation \(a^2 = b^2 + c^2\). Solving this equation for \(b\) we get: \(b^2 = a^2 - c^2\). Therefore, \(b^2 = 3^2 - 2^2 = 5\).
03

Write the Standard Form of the Equation

The standard form of the equation of an ellipse with a horizontal major axis and center at (h, k) is \((x - h)^2/a^2 + (y - k)^2/b^2 = 1\). Substituting in the values for \(h\), \(k\), \(a\), and \(b\) that were calculated earlier, the equation is \((x - 2)^2/3^2 + y^2/5 = 1\)

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Most popular questions from this chapter

Halley's comet has an elliptical orbit, with the sun at one focus. The eccentricity of the orbit is approximately 0.967 . The length of the major axis of the orbit is approximately 35.88 astronomical units. (An astronomical unit is about 93 million miles.) (a) Find an equation of the orbit. Place the center of the orbit at the origin, and place the major axis on the \(x\) -axis. (b) Use a graphing utility to graph the equation of the orbit. (c) Find the greatest (aphelion) and smallest (perihelion) distances from the sun's center to the comet's center.

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