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What is the equation of the line normal to # f(x)=1/(x^2-2) # at # x=1#?

Answer 1

Christopher Allers

#y = -(x + 2)/2#

Let's find the equation of the slope of the function.

So, #f'(x) = d/dxf(x) = d/dx(1/(x^2 - 2)) = (-1/(x^2 - 2)^2 * d/dx(x^2 - 2)) = (-2x)/(x^2 - 2)^2#

So, Slope of the Function #f(x)# at #x = 1# is

#f'(1) = (-2 * 1)/(1^2 - 2)^2 = (-2)/1 = -2#.

We know, The condition of perpendicularity of two straight lines in 2D Plane is,

#m_1m_2 = -1#, where #m_1# and #m_2# are the slopes of the straight lines respectively.

Let the Required Normal Line be #y = mx + c#.

According to the question,

#-2m = 1 rArr m = -1/2# [Condition of Perpendicularity]

So, The equation is now #y = -1/2x + c#...................(i)

Now, #f(x)# at #x = 1# :- #f(1) = 1/(1^2 - 2) = -1#

Let's put #y= -1# and #x = 1# in eq(i).

So, We get,

#color(white)(xxx)-1 = -1/2(1) + c#

#rArr -1 = -1/2 + c#

#rArr c = 1/2 - 1#

#rArr c = -1/2#

So, The final equation is :-

#color(white)(xxx)y = -1/2x - 1/2#

#rArr y = (-x - 2)/2#

#rArr y = -(x + 2)/2#

Hope this helps.

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Answer 2

Emilia Aguilar

The equation of the line normal to f(x)=1/(x^2-2) at x=1 is y = -2x + 3.

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