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.