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Chapter 7: Problem 43

Graph each ellipse and give the location of its foci. $$\frac{x^{2}}{25}+\frac{(y-2)^{2}}{36}=1$$

Short Answer

Expert verified
The graph of the given ellipse is centered at (0, 2), with its major axis along the y-axis of lengths 12 units and minor axis along the x-axis of lengths 10 units. The foci are at (0, 2- sqrt(11)) and (0, 2+ sqrt(11)).

Step by step solution

01

- Identify the centroid

In this equation, x is not translated and y is translated by 2 units upwards, so the centroid is at the coordinates (0, 2).
02

- Draw the axes and ellipse

The lengths of the semi axes are given by the square roots of the numbers under x and y. So, draw the major and minor axes of the ellipse according to the given lengths (6 units in the y direction and 5 units in the x direction). Then sketch the ellipse around these axes.
03

- Find the length of the focal distance (c)

Apply the formula c = sqrt(b^2 - a^2) . Here a = 5, b = 6, so we get c = sqrt( 36 - 25) = sqrt(11) units.
04

- Find the coordinates of the foci

An ellipse with a vertical major axis has its foci at (h, k 卤 c), so in this case, the coordinates of the two foci are (0, 2卤sqrt(11)).

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Most popular questions from this chapter

Describe how to graph \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Graph \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) in the same viewing rectangle for values of \(a^{2}\) and \(b^{2}\) of your choice. Describe the relationship between the two graphs.

Will help you prepare for the material covered in the next section. Divide both sides of \(4 x^{2}-9 y^{2}=36\) by 36 and simplify. How does the simplified equation differ from that of an ellipse?

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Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y+1)^{2}=-8 x$$
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