Question

What is the equation of the line that is normal to `f(x)= x/sqrt( x^2-2)` at `x=2`?

Answer 1

`y = sqrt2x - sqrt2`

Explanation:

I always prefer to rewrite my functions so I can simply use the Product Rule.

`f(x) = x(x^2-2)^-(1/2)`

Find the derivative to find the slope of the tangent line at `x=2`.

`f'(x) = (x^2-2)^-(1/2)-x/2(x^2-2)^(-3/2)(2x)`

`f'(x) = 1/sqrt(x^2-2)-x^2/(x^2-2)^(3/2)`

`f'(2) = 1/sqrt(2^2-2)-2^2/(2^2-2)^(3/2)`

`f'(2) = 1/sqrt(2)-4/(2)^(3/2) = -1/sqrt2= m_t`

To find the normal slope, we take the negative reciprocal of our tangent slope.

`m_n = sqrt2`

We can now construct an equation using a basic point-slope formula:

`y-y_o = m(x-x_o)`

`x_o` is given to us: `x_o = 2`

`y_o` can be solved by plugging in `x_o` back into our original equation:

`y_o = f(2) = (2)(2^2-2)^-(1/2) = 2/sqrt2`

`y - 2/sqrt2 = sqrt2(x-2)`

`y = sqrt2x-2sqrt2+2/sqrt2`

`y = sqrt2x - sqrt2`