Question

How do you find the equation of the tangent and normal line to the curve `y=1/x` at x=2?

Answer 1

Step 1 - Find the value of `y` when `x=2`:
`x=2=>y=1/2`

Step 2 - Find the gradient of the tangent at `x=2`:

`y=1/x`, `:. y=x^-1`
So, `(dy)/(dx)=(-1)x^-2=-1/x^2`

When `x=2=> (dy)/(dx)=-1/2^2=-1/4`

Step 3 - Eq'n of the Tangent when `x=2`:
The tangent passes through `(2,1/2)` and has gradient `-1/4`
Using `y-y_1=m(x-x_1)` we have the equation of the tangent is given by:
`y-1/2=-1/4(x-2)`
`:. 4y-2=-1(x-2)` (multiplying by 4)
`:. 4y-2=2-x`
`:. 4y+x=4`

Step 4 - Eq'n of the Normal when `x=2`:
Normal is perpendicular to Tangent, so
gradient of Normal = `-1/(-1/4)=4`

The normal passes through `(2,1/2)` and has gradient `4`
Using `y-y_1=m(x-x_1)` we have the equation of the normal is given by:
`y-1/2=4(x-2)`
`:. 2y-1=8(x-2)` (multiplying by 2)
`:. 2y-1=8x-16`
`:. 2y-8x=-15`