The equation of a curve is y=5-8/x. Show that the equation of the normal to the curve and the point P (2,1) is 2y+x=4

Answer 1

y = 5 - 8/x = 5 - 8(x^-1)

y' = 8(x^-2) = 8/x^2

for the point (2,1) y' = 8/2^2 = 8/4 = 2 = slope of the line tangent to the curve at that point. The normal will have the slope of a perpendicular line, with a negative inverse slope = -1/2

the equation of the normal line will be -1/2 = (y-1)/(x-2) or

y-1 = -x/2 +1

y = -x/2 +2

2y = -x + 4

2y +x = 4