Find the equation of the normal lines to the curve

Solution:

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

$y-2=-\frac{1}{3}(x-2)$

$\Rightarrow \quad 3 y-6=-x+2 \Rightarrow x+3 y=8$

at $(-2,-2) \quad y+2=-\frac{1}{3}(x+2)$

$\Rightarrow \quad 3 y+6=-x-2 \Rightarrow x+3 y=-8$

Therefore, the required equations are x + 3y = 8 and x + 3y = -8.