Given: equation of the curve 3x²–y² = 8, equation of line x+3y = 4

To find: the equation of the normal lines to the given curve which are parallel to the given line

Explanation:

Now given equation of curve as 3x² -y² = 8

Differentiating this with respect to x, we get

Now let slope of the normal to the curve be m₂ is given by

Substituting value from equation (i), we get

The given equation of the line is

x+3y = 4

⇒ 3y = 4-x

the slope of this line is

Since, slope of normal to the curve should be equal to the slope of the line which is parallel to the curve,

∴ m₂ = m₃

Substituting values from equation (ii) and (iii), we get

⇒ 3y = 3x

⇒ y = x……(iv)

Now substituting y = x in the equation of the curve, we get

3x² -y² = 8

⇒ 3x² -x² = 8

⇒ 2x² = 8

⇒ x² = 4

⇒ x = ±2

But from equation (iv)

y = x

⇒ y = ±2

Thus the points at which normal to the given curve is parallel to the given line are (2, 2) and (-2, -2)

Now the equations of the normal are given by

y-2 = m₂(x-2) and y+2 = m₂(x+2)

and

3y-6 = -x+2 and 3y+6 = -x-2

3y+x = 8 and 3y+x = -8

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