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: 62



Work Step by Step

$x^2+4y^2=5...(a)\\ 2x+8yy'=0\\ y'=-\frac{x}{4y}$ Equation of a tangent line at $(-5,0)$ $y-0=m(x-(-5))\\ y=m(x+5)\\ y=mx+5x\\ y=-\frac{x^2}{4y}-\frac{5x}{4y}\\ 4y^2=-x^2-5x\\ 4y^2+x^2=-5x$ From $(a)$ $5=-5x\\ x=-1\Longrightarrow y=1$ Tangent line at $(-1,1)$ $\frac{y-1}{0-1}=\frac{x-(-1)}{-5-(-1)}\\ -4y+4=-x-1\\ y=\frac{1}{4}x+\frac{5}{4}...(b)$ Plug in $x=3$ into $(b)$ $y=\frac{3}{4}+\frac{5}{4}\\ y=2$ Therefore, the lamp is located 2 units above the $x$-axis.
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