Question

Find the equations of the normals to the curve xy = 3 which are parallel to the

line 3x - y - 2 = 0.

Answer 1

y = 3x + 8.

y = 3x - 8

Explanation:

The slope of the tangent to the curve at a given point is obtained from the derivative.

xy = 3

y = 3/x

y = 3 x^-1

y' = -3x^-2

The slope of the normal (which is perpendicular to the tangent)

= -1 / -3x^-2 = 1/3 x^2.

Consider the given line:

3x - y - 2 = 0

y = 3x - 2 , so the slope of this line = 3.

Now this line is parallel to the normal so 1/3 x^2 = 3

Therefore x^2 = 9 and x = -3, 3.

Find the equations of the normals:

At x = -3, y = 3/-3 = -1, which gives the equation:

y - (-1) = 3(x - -3)

y = 3x + 8.

At x = 3 , y = 1

y - 1 = 3(x - 3)

y = 3x - 8.