Given that the normal to the curve 3x² – y² = 8 which is parallel to x + 3y = k.

Let (a, b) be the point of intersection of both the curve.

⇒ 3a² – b² = 8 ….(1)

and a + 3b = k ….(2)

Now, 3x² – y² = 8

On differentiating w.r.t. x,

Slope of the tangent at (a, b)

Slope of the normal at (a, b)

Slope of normal = Slope of the line

⇒ b = a ….(3)

Put (3) in (1),

3a² – a² = 8

⇒ 2a²=8

⇒ a = ±2

Case: 1

When a = 2, b = 2

⇒ x + 3y =k

⇒ k = 8

Case: 2

When a = -2, b = -2

⇒ x + 3y =k

⇒ k = -8

From both the cases,

The equation of the normal to the curve 3x² – y² = 8 which is parallel to x + 3y = k is x + 3y = ±8.