Question

The equation of the normal to the curve ay² = x³ at the point (am², am³) is ______.

Options

• 2y – 3mx + am³ = 0

• 2x + 3my 3am⁴ – am² = 0

• 2x + 3my + 3am⁴ – 2am = 0

• 2x + 3my – 3am⁴ – 2am² = 0

Answer 1

The equation of the normal to the curve ay² = x³ at the point (am², am³) is 2x + 3my – 3am⁴ – 2am² = 0.

Explanation:

We know that

Slope of tangents is `dy/dx`

Given

ay² = x³

Differentiating w.r.t.x

`(d(ay^2))/dx = (d(x^3))/dx`

`a(d(y^2))/dx = (d(x^3))/dx`

`a * (d(y^2))/dx xx dy/dy = 3x^2`

`a * 2y xx dy/dx = 3x^2`

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

Slope of tangent at (am², am³) is

`dy/dx|_((am^2"," am^3)) = (3(am^2)^2)/(2a(am^3))`

= `(3a^2m^4)/(2a^2m^3)`

= `3/2 m`

We know that

Slope of tangent × Slope of Normal = –1

`(3m)/2 xx "Slope of Normal" = –1`

Slope of Normal = `(-1)/((3m)/2)`

Slope of Normal = `(-2)/(3m)`

Finding equation of normal

We know that Equation of line at (x₁, y₁) and having Slope m is

y – y₁ = m(x – x₁)

Equation of Normal at (am², am³) and having Slope `(-2)/(3m)` is 

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

3m(y – am³) = –2(x – am²)

3my – 3am⁴ = –2x + 2am²

2x + 3my – 3am⁴ – 2am² = 0

2x + 3my – am²(3m² + 2) = 0

Required Equation of Normal is 2x + 3my – am²(3m² + 2) = 0