## Calculus 8th Edition

$2$
$x^2+4y^2=5...(a)\\ 2x+8yy'=0\\ y'=-\frac{x}{4y}$ 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.