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.
