# Family of orthogonal trajectories for a vertical parabola

1. Feb 15, 2007

### DaConfusion

1. The problem statement, all variables and given/known data. The correct answer is E

2. Relevant equations:Procedure from our text:

"Step 1. Determine the differential equation for the given family F(x, y,C) = 0.
Step 2. Replace y' in that equation by −1/y'; the resulting equation is the differential equation
for the family of orthogonal trajectories.
Step 3. Find the general solution of the new differential equation. This is the family of orthogonal
trajectories."

3. The attempt at a solution

Thank you all for looking/helping!

2. Feb 16, 2007

### HallsofIvy

Staff Emeritus
You might want to start by reversing x and y in you original equation: if a parabola has vertical axis and vertex at (3, 6) then it is of the form
y= C(x-3)2+ 6. Also, looking at the way the possible solutions are given, I would NOT multiply out the squares.

3. Feb 16, 2007

### DaConfusion

Oh...that vertical was confusing me like crazy! I kept thinking the axis is parallel to the asymptote not orthogonal. Thanks.

4. Feb 16, 2007

### DaConfusion

I still do not see it. Has to be some simple mistake...

5. Feb 16, 2007

### HallsofIvy

Staff Emeritus
At one point, you have
$$2(y-6)= y'(x+3)$$
(You have, accidently, "-" instead of "=")
but the next line is
$$\frac{2(y-6)}{x-3}= y'$$
where it obviously should be "x+3" instead of "x-3".

$$\frac{x^2}{2}+3x+ y^2- 12y= C$$
(Notice that I have changed your "-3x" to "+3x")
Complete the square:
that's the same as
$$x^2+ 6x+ 2(y^2- 12y) = 2C$$
$$x^2+ 6x+ 9+ 2(y^2- 12y+ 36)= 2C+ 9+ 72$$
$$= (x+ 3)^2+ 2(y- 6)^2= 2C+9+72= C'$$
which is one of your options.

By the way, it would be simpler to integrate
$$2\int (y-6)dy+ \int (x+3)dx= 0$$
as
[tex](y-6)^2+ \frac{1}{2}(x+3)^2= C[/itex]
This differs from your integral only by the constant C.

6. Feb 16, 2007

### DaConfusion

perfect!

Thanks for the help I can see it now.