Question

Find the equation of normal at the point `(am^2, am^3)` for the curve `ay^3 = x^3`?

Answer 1

The normal equation is:

`y = -2/(3m)x + am^3 + (2am)/(3)`

Explanation:

Firstly If:

`x = am^2 => x^3 = (am^2)^3 = a^3m^6`
`y = am^3 => y^2 = (am^3)^2 = a^2m^6`

Then equating `m^6` we get:

`x^3/a^3 = y^2/a^2 => ay^2 = x^3 \ \` and not `\ ay^3 = x^3`

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is `−1`).

Now that we have the correct cartesian equation we can determine an expression for the derivative by using the parametric or cartesian equation - both will yield identical;cal results:

Method 1 - Parametric Equations

We have, for some parameter `m` that

`x(m) = am^2`
`y(m) = am^3`

Differentiating wrt `m` we get

`dx/(dm) = 2am`
`dy/(dm) = 3am^2`

By the chain rule;

`dy/dx = (dy//dm)/(dx//dm)`
`" " = (3am^2)/(2am)`
`" " = (3m)/2`

Method 2 - Cartesian Equations

We have:

`ay^2 = x^3`

Differentiating implicitly:

`2aydy/dx = 3x^2`
`:. dy/dx = (3x^2)/(2ay)`

So, at the generic point `(am^2,am^3)`:

`dy/dx = (3(am^2)^2)/(2a(am^3))`
`" " = (3a^2m^4)/(2a^2m^3)`
`" " = (3m)/(2)`, as above

So, we have established that at the point `(am^2,am^3)` the gradient of the tangent, `m_T`, is given by the derivative#

`m_T = (3m)/(2)`

So, the gradient of the normal, `m_N`, is given by:

`m_N = -2/(3m)`

And, as the normal also passes through the same coordinate, we can find its equation using the point/slope form `y-y_1=m(x-x_1)`;

`y - (am^3) = -2/(3m)(x - am^2)`
`:. y - am^3 = -2/(3m)x + 2/(3m)am^2`
`:. y - am^3 = -2/(3m)x + (2am)/(3)`

`:. y = -2/(3m)x + am^3 + (2am)/(3)`