Given the equation of the curve:
[tex]y=5-\frac{8}{x}[/tex]We will differentiate the equation to find the slope of the tangential
So,
[tex]y^{\prime}=0-8\cdot-\frac{1}{x^2}=\frac{8}{x^2}[/tex]So, the value of the slope at the point P(2, 1) will be = m
[tex]m=\frac{8}{2^2}=\frac{8}{4}=2[/tex]The slope of the normal at the same point = m'
[tex]m^{\prime}=-\frac{1}{m}=-\frac{1}{2}[/tex]The equation of the line using the point-slope form will be:
[tex]\begin{gathered} (y-1)=-\frac{1}{2}(x-2) \\ y-1=-\frac{1}{2}x+1 \\ y+\frac{1}{2}x=2\rightarrow(\times2) \\ \\ 2y+x=4 \end{gathered}[/tex]So, the answer will be the equation of the normal line 2y + x = 4
B) The normal line meets the curve at point Q
We will find the point Q by substitution
Substitute with y from the given equation of the curve into the equation of the line:
[tex]\begin{gathered} 2(5-\frac{8}{x})+x=4 \\ 10-\frac{16}{x}+x=4\rightarrow(\times x) \\ \\ 10x-16+x^2=4x \\ x^2+6x-16=0 \\ (x+8)(x-2)=0 \\ x=-8,orx=2 \end{gathered}[/tex]Substitute with x into the equation of the curve to find y:
[tex]\begin{gathered} x=-8\rightarrow y=5-\frac{8}{-8}=6 \\ \\ x=2\rightarrow y=5-\frac{8}{2}=5-4=1 \end{gathered}[/tex]So, the answer will be
Q = (-8, 6) or Q = (2, 1)