Question

 Find the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4.

Answer 1

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.