Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 - Page 168: 57


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$x^2-xy+y^2=3\\ 2x-(xy'+y)+2yy'=0\\ 2x-xy'-y+2yy'=0\\ y'(2y-x)=y-2x\\ y'=\frac{y-2x}{2y-x}$ The ellipse crosses the $x$-axis when $y=0$ $x^2-xy+y^2=3\\ x^2=3\\ x=\pm\sqrt3$ Then we plug in $\pm\sqrt3$ into $x^2-xy+y^2=3$ and we get $(\sqrt3,0)$; $(-\sqrt3,0)$. Now we need to check if the tangents at those points are parallel. $(\sqrt3,0)$ $y'=\frac{y-2x}{2y-x}\\ y'=\frac{(0)-2(\sqrt3)}{2(0)-(\sqrt3)}\\ y'=2$ $(-\sqrt3,0)$ $y'=\frac{y-2x}{2y-x}\\ y'=\frac{(0)-2(-\sqrt3)}{2(0)-(-\sqrt3)}\\ y'=2$ Therefore, the tangent lines at $(\sqrt3,0)$ and $(-\sqrt3,0)$ are parallel as they both have the same gradient.
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