Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.3 The Ellipse - 10.3 Assess Your Understanding - Page 657: 49


Center: $(2,-1)$ Foci: $(1,-1),(3,-1)$ Vertices: $(2-\sqrt 3,-1),(2+\sqrt 3,-1)$ See graph

Work Step by Step

We are given the ellipse: $2x^2+3y^2-8x+6y+5=0$ Put the equation in standard form: $2(x^2-4x+4)-8+3(y^2+2y+1)-3+5=0$ $2(x-2)^2+3(y+1)^2=6$ $\dfrac{2(x-2)^2}{6}+\dfrac{3(y+1)^2}{6}=1$ $\dfrac{(x-2)^2}{3}+\dfrac{(y+1)^2}{2}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Identify $h,k,a,b$: $h=2$ $k=-1$ $a^2=3\Rightarrow a=\sqrt{3}$ $b^2=2\Rightarrow b=\sqrt {2}$ Determine $c$: $a^2=b^2+c^2$ $c^2=a^2-b^2$ $c^2=3-2$ $c^2=1$ $c=1$ Determine the center: $(h,k)=(2,-1)$ Determine the foci: $(h-c,k)=(2-1,-1)=(1,-1)$ $(h+c,k)=(2+1,-1)=(3,-1)$ Determine the vertices: $(h-a,k)=(2-\sqrt 3,-1)$ $(h+a,k)=(2+\sqrt 3,-1)$ Determine the $y$-intercepts: $(h,k-b)=(2,-1-\sqrt 2)$ $(h,k+b)=(2,-1+\sqrt 2)$ Graph the ellipse:
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