Question

What is the equation of the line normal to `f(x)= -1/(5-x^2)` at `x=-3`?

Answer 1

Equation of line normal is `32x+12y+93=0`

Explanation:

Slope of the tangent of a curve, at a given point, is given by the value of the first derivative at that point. As normal is perpendicular to tangent, if `m` is the slope of tangent, slope of normal would be `-1/m`.

To find the point, let us put `x=-3` in `f(x)=-1/(5-x^2)` and `f(-3)=-1/(5-(-3)^2)=-1/-4=1/4`.

Hence, we are seeking normal at `(-3,1/4)`

`(df)/(dx)=-(-1/(5-x^2)^2)xx(-2x)=-(2x)/(5-x^2)^2`

Hence slope of tangent at `x=-3` is

`-(2(-3))/(5-(-3)^2)^2=6/(-4)^2=6/16=3/8`

and slope of normal is `-1/(3/8)=-8/3`

and equation of line normal is

`(y-1/4)=-8/3(x-(-3))`

or `12y-3=-32x-96` or `32x+12y+93=0`