Answer:
y = 3x + 8.
y = 3x - 8
Step-by-step 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.
