Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 36

Answer

Ellipse: - the vertices are $(3\pm \sqrt 8,4)$, $(3,4\pm \sqrt 6)$ - the foci are $(3\pm \sqrt{2},4)$ - the center is $(3,4)$

Work Step by Step

By completing the square, we can write the equation $$ 8 y^{2}+6 x^{2}-36 x-64 y+134=0 $$ in the form $$ 8 (y-4)^2+6 (x-3 )^2=48. $$ Then we get $$ \left(\frac{x-3}{ \sqrt 8}\right)^{2}+\left(\frac{y-4}{\sqrt 6}\right)^{2}=1 $$ which is an ellipse with $a=\sqrt 8, b=\sqrt 6$ and hence $ c=\sqrt{a^2-b^2}=\sqrt{2} $ So, we have: - the vertices are $(3\pm \sqrt 8,4)$, $(3,4\pm \sqrt 6)$ - the foci are $(3\pm \sqrt{2},4)$ - the center is $(3,4)$
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