What is the equation of the line normal to f(x)=1/(x^2-2) f(x)=1x22 at x=1x=1?

1 Answer

y = -(x + 2)/2y=x+22

Explanation:

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.