Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1077: 114

Answer

See below:
1570771742

Work Step by Step

The larger number is under the expression involving $y$, so the major axis is vertical and parallel to the $y$ -axis. Compare equation (II) $\frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1$ with the standard form $\frac{{{\left( x-h \right)}^{2}}}{{{b}^{2}}}+\frac{{{\left( y-k \right)}^{2}}}{{{a}^{2}}}=1$ , $h=3$, $k=-3$ $\begin{align} & {{a}^{2}}=36\Rightarrow a=\pm 6 \\ & {{b}^{2}}=9\Rightarrow b=\pm 3 \end{align}$ So, the center of the ellipse is given by $\left( h,k \right)=\left( 3,-3 \right)$. For a vertical major axis with center $\left( 3,-3 \right)$, the vertices are given by points $\begin{align} & \left( 3,-3+6 \right)=\left( 3,3 \right) \\ & \left( 3,-3-6 \right)=\left( 3,-9 \right) \\ \end{align}$ End-points of the minor axis are given by the points $\begin{align} & \left( 3+3,-3 \right)=\left( 6,-3 \right) \\ & \left( 3-3,-3 \right)=\left( 0,-3 \right) \\ \end{align}$ The coordinates of the foci of an ellipse are given by $\left( h+c,k-c \right)$ We know that, $\begin{align} & {{c}^{2}}={{a}^{2}}-{{b}^{2}} \\ & =36-9 \\ & =27 \\ & c=\pm 3\sqrt{3} \end{align}$ For a vertical major axis with center $\left( 3,-3 \right)$, the foci are at points $\left( 3,-3+3\sqrt{3} \right)$ And $\left( 3,-3-3\sqrt{3} \right)$ Consider the equation of an ellipse: $4{{x}^{2}}+{{y}^{2}}-24x+6y+9=0$ (I) Now, we will convert the equation in standard form as follows: $\begin{align} & 4{{x}^{2}}+{{y}^{2}}-24x+6y+9=0 \\ & 4{{x}^{2}}-24x+{{y}^{2}}+6y+9=0 \\ & 4\left( {{x}^{2}}-6x \right)+\left( {{y}^{2}}+6y \right)+9=0 \\ & 4\left( {{x}^{2}}-6x+9 \right)-36+\left( {{y}^{2}}+6y+9 \right)-9+9=0 \end{align}$ It can be further simplified as: $\begin{align} & 4{{\left( x-3 \right)}^{2}}+{{\left( y+3 \right)}^{2}}-36-9+9=0 \\ & 4{{\left( x-3 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=36 \\ & \frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1 \end{align}$ So one will have the equation of the ellipse as: $\frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1$ (II) Therefore, the foci of the ellipse are $\left( 3,-3+3\sqrt{3} \right)$ and $\left( 3,-3-3\sqrt{3} \right)$.
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