Question

Find the equations of the normals to the curve 3x² – y² = 8, which are parallel to the line x + 3y = 4.

Answer 1

Let P(x₁, y₁) be the foot of the required normal to the curve 3x² – y² = 8.

Differentiating 3x² – y² = 8 w.r.t. x, we get

`3 xx 2x - 2ydy/dx = 0`

∴ `-2ydy/dx` = – 6x

∴ `dy/dx = (3x)/y`

∴ `(dy/dx)_("at" (x_1, y_1)` = `(3x_1)/(y_1)` = slope of the tangent at (x₁, y₁)

∴ slope of the normal at P(x₁, y₁)

= `m_1 = (-1)/((dy/dx)_("at" (x_1, y1)`

= `-(y_1)/(3x_1)`

The slope of the line x + 3y = 4 is `m_2 = (-1)/(3)`

Since, the normal at P(x₁, y₁) is parallel to the line
x + 3y = 4, m₁ = m₂

∴ `-(y_1)/(3x_1) = - (1)/(3)`  ∴ y₁ = x₁

Since, (x₁, y₁) lies on the curve 3x² – y² = 8,
3x₁² – y₁² = 8
∴ 3x₁² – x₁² = 8           ...[∵ y₁ = x₁]
∴ 2x₁² = 8
∴ x₁² = 4
∴ x₁ = ± 2
When x₁ = 2, y₁ = 2
When x₁ = – 2, y₁ = – 2
∴ the  coordinate of the point P are (2, 2) or (– 2, – 2) and the slop of the normal is m₁ = m₂ = ` - (1)/(3)`

∴ the equation of the normaal at (2, 2) is

`y - 2 = -(1)/(3)(x - 2)`

∴ 3y – 6 = – x + 2
∴ x + 3y – 8 = 0
and the equation of the normal at (– 2, – 2) is

`y + 2 = -(1)/(3)(x + 2)`

∴ 3y – 6 = – x + 2
∴ x + 3y + 8 = 0
Hence, the equations of the normals are
x + 3y – 8 = 0 and x + 3y + 8 = 0.