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.

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

452,489 to the nearest thousand is 452,000