Answer
$2$
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.
