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

Question

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

Scarlett Allison · Accepted Answer

#y = sqrt2x - sqrt2#

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#

Emma Aldridge · Answer

undefined To find the equation of the line that is normal to the function f(x) = x/√(x^2 - 2) at x = 2, we need to determine the slope of the tangent line at x = 2 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [√(x^2 - 2) - x(1/2)(2x)] / (x^2 - 2)^(3/2)
       = [√(x^2 - 2) - x^2] / (x^2 - 2)^(3/2)

Next, we substitute x = 2 into f'(x) to find the slope of the tangent line at x = 2:

f'(2) = [√(2^2 - 2) - 2^2] / (2^2 - 2)^(3/2)
      = [√2 - 4] / (2 - 2)^(3/2)
      = [√2 - 4] / 0

Since the denominator is zero, the slope is undefined. This means that the tangent line is vertical, and the normal line will be horizontal.

Therefore, the equation of the line that is normal to f(x) = x/√(x^2 - 2) at x = 2 is simply the equation of the horizontal line passing through the point (2, f(2)).

To find the y-coordinate of this point, we substitute x = 2 into f(x):

f(2) = 2 / √(2^2 - 2)
     = 2 / √2
     = 2√2 / 2
     = √2

Thus, the equation of the line that is normal to f(x) = x/√(x^2 - 2) at x = 2 is y = √2.