# Derivatives and parallel lines

1. Dec 5, 2007

### chris40256

1. The problem statement, all variables and given/known data

Consider the curve defined by x^2+xy +y^2=27

a. Write an expression for the slope of the curve at any point (x,y)
b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leads to your conclusion.
c. Find the points on the curve where the lines tangent to the curve are vertical.

3. The attempt at a solution

a. I used implicit differentiation
x^2 +xy +y^2 =27
2x +y + xy' +2yy' = 0
y'(x+2y)= - (2x+y)
y' = -(2x+y)/(x+2y). Im pretty sure this is right.

b. i know that the lines must have the same slope for them to be parallel, but im not actually sure what else to do with this. how do u find the x-intercepts and show that their slopes are equal?

c. the vertical line would be like x=k, then you plug k into the orginal curve and then solve for k? i got 6 and -6 is that right? can someone please help me with this

2. Dec 5, 2007

### Shooting Star

a. Correct

b. Find x-intercepts of the curve by putting y=0, and find the x values. Then put these x and y in dy/dx.

c. Find where dx/dy is zero.

3. Dec 5, 2007

### Office_Shredder

Staff Emeritus
(a) is almost right, but you neglect the possibility that x+2y=0. You would either have to find an alternative expression for that case, or show it never happens on the curve (e.g. by substituting x=-2y into the original equation)

For (b), I suggest being really slick and not calculating where the x-intercepts are located if you can (try plugging y=0 into your dy/dx expression. Where can this go wrong?). Note a line that's tangent to a curve has a slope equal to dy/dx

For (c), if the curve is parallel to a vertical line, then dy/dx is +/- infinity. Taking a page from part (a), when does this occur?

4. Dec 11, 2007

### Shooting Star

I didn't quite get what exactly you meant.

Last edited by a moderator: Dec 11, 2007
5. Dec 11, 2007

### HallsofIvy

Staff Emeritus
He is saying, rather than set y= 0 in the original equation, to find the x-intercepts, set y= 0 in the formula for the derivative. It happens, for this particular problem, that you don't need to know x!