Question

The equation of a curve is y = 5-8/xa) Show that the equation of the normal to the curve at thepoint P(2,1) is 2y + x = 4.(TIP: Use point-slope form!)b) This normal meets the curve again at the point Q.Find the coordinates of Q. (TIP: use substitution!)

Answer 1

Given the equation of the curve:

`y=5-(8)/(x)`

We will differentiate the equation to find the slope of the tangential

So,

`y^(\prime)=0-8\cdot-(1)/(x^2)=(8)/(x^2)`

So, the value of the slope at the point P(2, 1) will be = m

`m=(8)/(2^2)=(8)/(4)=2`

The slope of the normal at the same point = m'

`m^(\prime)=-(1)/(m)=-(1)/(2)`

The equation of the line using the point-slope form will be:

`\begin{gathered} (y-1)=-(1)/(2)(x-2) \\ y-1=-(1)/(2)x+1 \\ y+(1)/(2)x=2\rightarrow(*2) \\ \\ 2y+x=4 \end{gathered}`

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:

`\begin{gathered} 2(5-(8)/(x))+x=4 \\ 10-(16)/(x)+x=4\rightarrow(* x) \\ \\ 10x-16+x^2=4x \\ x^2+6x-16=0 \\ (x+8)(x-2)=0 \\ x=-8,orx=2 \end{gathered}`

Substitute with x into the equation of the curve to find y:

`\begin{gathered} x=-8\rightarrow y=5-(8)/(-8)=6 \\ \\ x=2\rightarrow y=5-(8)/(2)=5-4=1 \end{gathered}`

So, the answer will be

Q = (-8, 6) or Q = (2, 1)