Find the equations of the normals to the curve xy = 3 which are parallel to the line 3x - y - 2 = 0.

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.

connie < felicia < stew

step-by-step explanation:

stew:

we need to find the slope

m = (y2-y1)/ (x2-x1)

    = (2650-795)/(10-3)

    = 1855/7

265 words per minute

connie:

we need to find the slope

m = (y2-y1)/ (x2-x1)

we have two points (0,0) and (4,800)

m = (800-0)/(4-0)

  m = 800/4

    200 words per minute

felicia:

260 words per minute

200< 260 < 265

connie < felicia < stew