Question

Find the equations of tangent and normal to the curve y = 6 - x² where the normal is parallel to the line x - 4y + 3 = 0

Answer 1

Let P (x₁, y₁) be the point on the curve y = 6 - x², where the normal is parallel to the line x - 4y + 3 = 0

Differentiating  y = 6 - x² w.r.t. x, we get

`dy/dx=d/dx(6-x^2)=0-2x=-2x`

∴`(dy/dx)_(at(x_1,y_1)) =-2x_1`

= slope of the tangent at (x₁, y₁)

∴ slope of the normal at (x₁, y₁) = `1/(2x_1)`

Let `m_1=1/(2x_1)`

The slope of the line x - 4y + 3 = 0 is m₂ `=-1/(-4)=1/4`

Since the normal at (x₁, y₁) is parallel to the line x - 4y + 3 = 0, m₁m₂

∴ `1/(2x_1)=1/4`

∴ 2x₁ = 4

∴ x₁ = 2

Since (x₁, y₁) lies on the curve y = 6 - x², y₁ = 6 - x₁²

∴ y₁ = 6 - (2)² = 2      ...[∵ x₁ = 2]

∴ the coordinates of the point are (2, 2) and slope of the normal = m₁ = m₂ = `1/4`

∴ the slope of the tangent = `-1/(m_1) = -4`

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

y - 2 = -4(x -2)

∴  y - 2 = -4x + 8

∴  4x + y = 10

and the equation of the normal at (2, 2) is

`y - 2 = 1/4 (x-2)`

∴ 4y - 8 = x - 2

∴ x - 4y + 6 = 0

Hence, equations of tangent and normal are 4x + y = 10 and x - 4y + 6 = 0 respectively