Given curve, 3x2 – y2 = 8
Differentiating both sides w.r.t. x, we get
6x – 2y. dy/dx = 0 ⇒ -2y(dy/dx) = -6x ⇒ dy/dx = 3x/y
So, slope of the tangent to the given curve = 3x/y
Thus, the normal to the curve = -1/(3x/y) = -y/3x
Now, differentiating both sides of the given line x + 3y = 4, we have
1 + 3.(dy/dx) = 0
dy/dx = -1/3
As the normal to the curve is parallel to the given line x + 3y = 4
We have, -y/3x = -1/3 ⇒ y = x
On putting the value of y in 3x2 – y2 = 8, we get
3x2 – x2 = 8
2x2 = 8 ⇒ x2 = 4 ⇒ x = ±2
So, y = ±2
Thus, the points on the curve are (2, 2) and (-2, -2).
Now, the equation of the normal to the curve at (2, 2) is given by
Therefore, the required equations are x + 3y = 8 and x + 3y = -8.