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

1 Answer
Oct 22, 2017

#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#