Answered: Ellipse: The major axis is horizonal…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

Ellipse: The major axis is horizonal with length of 8, the minor axis is length 4 and the center is ( − 1 , − 3 ).

State the eccentricity of this ellipse rounded to 4 decimal places.

Write the standard rectangular form of equation for this ellipse.

C. Sketch the ellipse. Label the vertices, co-vertices, foci and center

Expert Solution

Given,

Center of the ellipse is $(h, k) = (- 1, - 3)$ .

Length of major axis is 8. That is,

$2 a = 8 \Rightarrow a = 4$

Length of minor axis is 4. That is,

$2 b = 4 \Rightarrow b = 2$

Now by using the relationship $a^{2} = b^{2} + c^{2}$ , we get

$4^{2} = 2^{2} + c^{2} \Rightarrow c^{2} = 12 \Rightarrow c = \sqrt{12}$

Now eccentricity of the ellipse is,

$e = \frac{c}{a} \Rightarrow e = \frac{\sqrt{12}}{4} \Rightarrow e = 0.8660$

Step 2

Equation of the ellipse is,

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 \Rightarrow \frac{{(x - (- 1))}^{2}}{4^{2}} + \frac{{(y - (- 3))}^{2}}{2^{2}} = 1 \Rightarrow \frac{{(x + 1)}^{2}}{16} + \frac{{(y + 3)}^{2}}{4} = 1 \Rightarrow {(x + 1)}^{2} + 4 {(y + 3)}^{2} = 16 \Rightarrow x^{2} + 2 x + 1 + 4 (y^{2} + 6 y + 9) = 16 \Rightarrow x^{2} + 2 x + 1 + 4 y^{2} + 24 y + 36 = 16 \Rightarrow x^{2} + 4 y^{2} + 2 x + 24 y + 21 = 0$

Foci are,

$(h - c, k) & (h + c, k) \Rightarrow (- 1 - \sqrt{12}, - 3) & (- 1 + \sqrt{12}, - 3) \Rightarrow (- 4.464, - 3) & (2.464, - 3)$

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