Question

find the equation of tangent normal to the curve y=3x-x^3 at points where it cuts the x-axis​

Answer 1

Correct Question:-

Find the equations of lines normal to the curve $\rm{y=3x-x^3}$ at points where it cuts the x axis.

$\quad$

Solution:-

The given curve,

$\longrightarrow\rm{y=3x-x^3}$

Differentiating wrt x we get,

$\longrightarrow\rm{\dfrac{dy}{dx}=3-3x^2}$

This term is slope of tangent on the curve at a point (x, y).

Taking reciprocal,

$\longrightarrow\rm{\dfrac{dx}{dy}=\dfrac{1}{3-3x^2}}$

and multiplying by -1,

$\longrightarrow\rm{-\dfrac{dx}{dy}=\dfrac{1}{3x^2-3}}$

Now this term is slope of normal on the curve at a point (x, y).

$\longrightarrow\rm{n(x)=\dfrac{1}{3(x^2-1)}}$

$\quad$

The points where the curve $\rm{y=3x-x^3}$ cuts x axis is obtained by equating y = 0.

$\longrightarrow\rm{0=y}$

$\longrightarrow\rm{0=3x-x^3}$

$\longrightarrow\rm{0=x(3-x^2)}$

$\longrightarrow\rm{0=x(\sqrt3-x)(\sqrt3+x)}$

$\Longrightarrow\rm{x\in\left\{-\sqrt3,\ 0,\ \sqrt3\right\}}$

So the points where the curve cuts x axis are (-√3, 0), (0, 0) and (√3, 0).

$\quad$

Consider the point (-√3, 0).

The slope of normal at this point is,

$\longrightarrow\rm{n(-\sqrt3)=\dfrac{1}{3((-\sqrt3)^2-1)}}$

$\longrightarrow\rm{n(-\sqrt3)=\dfrac{1}{6}}$

Then equation of normal at this point is,

$\longrightarrow\rm{y-0=\dfrac{1}{6}(x+\sqrt3)}$

$\longrightarrow\bf{x-6y+\sqrt3=0}$

$\quad$

Consider the point (0, 0).

The slope of normal at this point is,

$\longrightarrow\rm{n(0)=\dfrac{1}{3((0)^2-1)}}$

$\longrightarrow\rm{n(0)=-\dfrac{1}{3}}$

Then equation of normal at this point is,

$\longrightarrow\rm{y-0=-\dfrac{1}{3}(x-0)}$

$\longrightarrow\bf{x+3y=0}$

$\quad$

Consider the point (√3, 0).

The slope of normal at this point is,

$\longrightarrow\rm{n(\sqrt3)=\dfrac{1}{3((\sqrt3)^2-1)}}$

$\longrightarrow\rm{n(\sqrt3)=\dfrac{1}{6}}$

Then equation of normal at this point is,

$\longrightarrow\rm{y-0=\dfrac{1}{6}(x-\sqrt3)}$

$\longrightarrow\bf{x-6y-\sqrt3=0}$

$\quad$

Hence the equations of lines normal to the curve $\rm{y=3x-x^3}$ at points where it cuts x axis are,

$\longrightarrow\left\{\begin{array}{ll}\bf{x-6y-\sqrt3}&\bf{=0}\\\bf{x+3y}&\bf{=0}\\\bf{x-6y+\sqrt3}&\bf{=0}\end{array}\right.$