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Chapter 7: Problem 26

Find the standard form of the equation of each ellipse satisfying the given conditions. $$\text { Foci: }(-2,0),(2,0) ; \text { vertices: }(-6,0),(6,0)$$

Short Answer

Expert verified
The standard form of the equation for the ellipse is \(x^2/36 + y^2/32 = 1\).

Step by step solution

01

Find the centre's coordinates.

The centre of the ellipse is the midpoint of the foci or vertices. Therefore, the midpoint of foci (-2,0) and (2,0) or vertices (-6,0) and (6,0) would yield the same result. Using the midpoint formula, the centre's coordinates (h,k) are: \(h = (x_1 + x_2)/2\), \(k = (y_1 + y_2)/2\). Plugging in the coordinates of the foci or vertices gives \(h = (2-2)/2 = 0\) and \(k = (0+0)/2 = 0\), so the centre is at (0,0).
02

Calculate the values of a, b, and c.

The distance between the foci divided by 2 is the value of c. Here, c = |(-2) - 2|/2 = 2. The distance between the vertices divided by 2 provides the value a. Here, a = |(-6) - 6|/2 = 6. The relation between a, b, and c in an ellipse is \(a^2 = b^2 + c^2\). Substituting a and c, we get \(b^2 = a^2 - c^2 = 6^2 - 2^2 = 32\). Therefore, b = \(\sqrt{32}\).
03

Write the standard form of the equation.

The standard form of the equation for a horizontal ellipse is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). By substituting h = 0, k = 0, a = 6, b = \(\sqrt{32}\), we get the standard form of the equation for the ellipse as \(x^2/36 + y^2/32 = 1\).

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Most popular questions from this chapter

Find the standard form of the equation of each ellipse satisfying the given conditions. Major axis vertical with length \(10 ;\) length of minor axis \(=4 ;\) center: \((-2,3)\)

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$9 x^{2}+16 y^{2}-18 x+64 y-71=0$$

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Graph each ellipse and give the location of its foci. $$\frac{(x+1)^{2}}{2}+\frac{(y-3)^{2}}{5}=1$$
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