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{(dy)/(dx)=3-3x^2}`

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

Taking reciprocal,

`\longrightarrow\rm{(dx)/(dy)=(1)/(3-3x^2)}`

and multiplying by -1,

`\longrightarrow\rm{-(dx)/(dy)=(1)/(3x^2-3)}`

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

`\longrightarrow\rm{n(x)=(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)=(1)/(3((-\sqrt3)^2-1))}`

`\longrightarrow\rm{n(-\sqrt3)=(1)/(6)}`

Then equation of normal at this point is,

`\longrightarrow\rm{y-0=(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)=(1)/(3((0)^2-1))}`

`\longrightarrow\rm{n(0)=-(1)/(3)}`

Then equation of normal at this point is,

`\longrightarrow\rm{y-0=-(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)=(1)/(3((\sqrt3)^2-1))}`

`\longrightarrow\rm{n(\sqrt3)=(1)/(6)}`

Then equation of normal at this point is,

`\longrightarrow\rm{y-0=(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.`