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College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter7: Conic Sections And Quadratic Systems

Section7.CT: Chapter Test

Problem 9CT

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Hi! I'm trying to figure out how to write an equation in the standard form of the ellipse. I'm so confused at this concept and everywhere I look keeps giving me a different answer on how to do it.

center at (-4, 5), c/a = 1/3, vertex at (-4, -4)

Expert Solution

Step 1

Given that

Center of the ellipse is (-4,5),

$\frac{c}{a} = \frac{1}{3}$ ,

One vertex of the ellipse is (-4,-4)

From the x-cordinate in the vertex and center, it is clear that the major axis is parallel to y-axis.

The standard form of an ellipse is give by

$\frac{{(x - h)}^{2}}{b^{2}} + \frac{{(y - k)}^{2}}{a^{2}} = 1 \to 1$

Center $(h, k) = (- 4, 5)$ ,

Vertex $(h, k - a) = (- 4, - 4)$

$h = - 4, k = 5 k - a = - 4 5 - a = - 4 5 + 4 = a a = 9$

Also,

$\frac{c}{a} = \frac{1}{3} \frac{c}{9} = \frac{1}{3} c = \frac{9}{3} c = 3$

To find b:

$c = \sqrt{a^{2}} - b^{2} c^{2} = a^{2} - b^{2} 3^{2} = 9^{2} - b^{2} b^{2} = 9^{2} - 3^{2} = 81 - 9 b^{2} = 72$

Foci $(h, k \pm c) = (- 4, 5 \pm 3) = (- 4, 8), (- 4, 2)$

Vertices $(h, k \pm a) = (- 4, 5 \pm 9) = (- 4, - 4), (- 4, 14)$

Covertices $(h \pm b, k) = (- 4 \pm 8.49, 5) = (4.5, 5) (- 12.5, 5)$

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